Csir Net Solution 2011 2017 PDF
Csir Net Solution 2011 2017 PDF
Csir Net Solution 2011 2017 PDF
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
NET December 2017 (Solution)
PART A
Q1. A leaf appears green in daylight. If this leaf were observed in red light, what colour
would it appear to have?
(a) Green (b) Black-Brown (c) Red (d) Blue
Ans. : (b)
Solution: We know that no object have their own colour, the colour reflected by any object
appears its own colour. An object which is black in colour does not reflect any colour of
the sun light.
So, the leaf which appears green in day light will be observed in red light would be black
in colour.
Q2. The distance from Nehrunagar to Gandhinagar is 27 km . A and B start walking from
Nehrunagar towards Gandhinagar at speeds of 5 km / hr and 7 km / hr , respectively. B
reaches Gandhinagar, returns immediately, and meets A at Indiranagar. What is the
distance between Nehrunagar and Indiranagar? (Assume all three cities to be in one
straight line)
(a) 12.5 km (b) 22.5 km (c) 4.5 km (d) 13.5 km
Ans. (b)
27 km Gandhinagar
Solution:
27 x km x km
Nehrungar
Total distance covered by B to reach Indiranagar after reaching Gandhinagar from
Nehrunagar 27 km x km
27 x
The time taken by B to cover this distance: hour
7
27 x
The time taken by A to reach Indiranagar from Nehrunagar hour
5
According to question:
27 x 27 x
, or 135 5 x 189 7 x
7 5
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54
or, 12 x 189 135 x 4.5 km
12
Hence, the distance between Nehrunagar and Indiranagar is 27 4.5 22.5 km
Q3. A sphere G of radius b is fixed mid-air and several spheres identical to the first one are
shot at it with their velocities parallel to each other. If the shot spheres fall within an
imaginary cylinder of radius a b a , then the fraction of spheres that will hit G is
Ans.: (b)
Solution: The area of sphere G of radius b 4 b 2
Let, the number of sphere shot, which is similar to sphere G be N , then total
Area of N space N .4 b 2
As, the shot sphere fall in an imaginary cylinder of radius a , then
Area of cylinder a 2
The fraction of spheres that will hit
Area of total sphere falling in cylinder N 4 b 2
G
N .Area of cylinder N a2
4b 2
Fraction
a2
Q4. Five persons A, B, C , D and E are sifting in a row with C in the middle of the group. If
D is at an extreme end and there are at least two persons between B and E , then which
of the following statements is incorrect?
(a) E can be on extreme left (b) E can be on extreme right
(c) A cannot be on extreme left (d) A is always a neighbour of B or D
Ans. : (d)
Solution: According to the question, there may be two cases, which are:
Case-1 D B C A E
E A C B D
Case -2
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E would be either in extreme left or extreme right position. A cannot be on extreme left.
A is always neighbour of E and C .
Q5. In a group of students, 30 % play only cricket, 20 % play only football and 10 % play
only basketball. 20 % of the students play both football and cricket, 15 % play both
basketball and cricket, 10 % play both football and basketball. 15 students play no
games, while 5 % of the students play all three games. What is the total number of
students?
(a) 300 (b) 250 (c) 350 (d) 400
Ans. : (a)
Solution: Total percentage of students, who likes games:
10% 30% 20% 5% 5% 10% 15% 95%
Hence the number of students who do not participate in the games 5% of the total
number of students 10% 10% 30% Cricket
Basketball
Total number of students 5%
5% 15%
15 100 Football
300 20%
5
Q6. When Ramesh was at the age of 8 years, he hammered a nail into a large tree to mark his
height. If the tree grows 2 cm/year, how much higher would the nail be after 5 years?
(a) 5cm higher (b) 0 cm higher (c) 10 cm higher (d) 8cm higher
Ans. (b)
Solution: Even if tree grows upwards, the position of nail will remain same after
5 years Nail Tree
Q7. Find the missing number
17 15 13 12 25 24 41 40
8 5 7 ?
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Solution: The number system is based on following patterns:-
17 2 152 17 15 17 15 32 2 64 82
132 122 13 12 13 12 25 1 25 52
252 242 25 24 25 24 49 1 49 7 2
412 402 41 40 41 40 81 1 81 92
Q8.
A
B won
lost
C
10 20 30 40 50 60 70 80 90 100110120130140150
Number of seats
The bar chart above shows number of seats won by four political parties A, B, C and D .
Which party won the largest proportion of seats it contested?
(a) A (b) B (c) C (d) D
Ans.: (b)
Solution: The proportion of seats of party A , out of total number of seats contested by it
100
150
The proportion of seats of party B , out of total number of seats contested by it
56 8 120
70 10 150
The proportion of seats of party C , out of total number of seats contested by it
72 18 108
100 25 150
The proportion of seats of party D , out of total number of seats contested by it
20 1 75
40 2 150
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B wins 120 seats if she will fight on 150 seats, hence, the largest proportion of seats
was won by B
Q9. The molar fraction of hydrochloric acid in an extremely dilute’ aqueous solution is
doubled. The pH of the resulting solution is
(a) approximately doubled (b) approximately halved
(c) increased (d) reduced
Ans.: (d)
Solution: Since PH is defined by the formula PH log H
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Solution: Number of vowels 5
Number of consonant 21
Here, our condition is to take atleast one consonant, but not consecutive, so we are left
with two cases-
Case I: Only one consonant-
Number of ways 3 21 5 5 1575
21 5 4 3 1260
Case II: Two consonant, but at alternate position,
Number of ways 21 21 5 2205
Therefore, total number of ways 1575 2205 3780
Q12. Which one of the following graphs represents f x sin x cos x ?
1 0.5
(a) 0.5 (b)
0 0
0.5
1 0.5
0 /2 3 / 2 2 0 /2 3 / 2 2
1 0.5
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Q13. There are two gas parcels of equal volume, A and B at the same temperature and
pressure. Parcel A is one mole of water vapour, while parcel B is one mole of dry air.
Which of the following is TRUE?
(a) Parcel A is heavier then Parcel B
(b) Parcel B is heavier than Parcel A
(c) Both parcels are equally heavy
(d) Without temperature and pressure data, their relative masses cannot be determined
Ans.: (b)
Solution: Water vapour content H 2O
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Q15. Find the next pattern in the following sequence:
(a) (b)
(c) (d)
Ans.: (c)
Solution: The arrow of first part of the figure changed alternately. The point on rectangular part
follow the series of 1, 2, 2, 4 and changes its position alternately.
The square under triangle comes out and it will be covered by triangle
Q16. DRQP is a small square of side a in the corner of a big square D R C
ABCD of side A. What is the ratio of the area of the quadrilateral P Q
PBRQ to that of the square ABCD , given A / a 3 ?
(a) 2 / 9 (b) 1/ 6
(c) 1/ 3 (d) 2 / 7 A B
Ans. : (a)
Solution: According to the question, let the side of D xR C
x x
AB 3 x BC CD AD and the side of P x Q
DP PQ RQ DR x 3x 3x
2x
The area of the quadrilateral PBRQ The area of square
A B
ABCD Area of square PDRQ Area of right angle triangle
ABP Area of Right angle triangle BCR .
1 1
3x 3x x x 3x 2 x 3x 2 x
2 2
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9 x x 3x 3x 2 x
2 2 2 2 2
1 m vertical portion and a 1 m portion tilted at 450 to the vertical. What will be the area
of wire mesh required?
(a) 200 m 2 (b) 241.4 m 2 (c) 400 m 2 (d) 170.7 m 2
Ans.: (a)
Solution: The area of the fence lying on tilted portion 100 1 100 m 2
1m
100 m
Q18. The average staff salary of a company is Rs. 8000 / - . A new guard and a new manager
are recruited with salaries of Rs. 5000 / - and 20000 / - , respectively. What is the current
staff strength if the new average salary is Rs. 4000 / - more than that of the guard?
(a) 7 (b) 9 (c) 10 (d) 11
Ans.: (b)
Solution: Let the current staff strength be x . Then the total salary of the staff Rs. 8000 x .
After a new managers, the sum of the salary of the staff is new
Rs.8000 x Rs.5000 Rs.20000
Rs.8000 x Rs.25000
The new average of the salary of the staff
Rs.8000 x Rs.25000
x
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According to the question
Rs.8000 x Rs.25000
Rs.5000 Rs.4000
x2
or, 8000 x 25000 9000 x 18000
or, 1000 x 7000
x7
And hence, strength of current staff 7 2 9
Thus option (b) is correct.
Q19. A bird flies along the three sides of a field in the shape of an equilateral triangle at speeds
of 2, 4,8 km / hr , respectively. The average speed of the bird is
24 14 22
(a) km / hr (b) km / hr (c) km / hr (d) 4 km / hr
7 3 7
Ans.: (a)
Solution: Let the side of a field of in the shape of an equilateral triangle x km
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4x
A pay Rs. for n copies of books. While B after getting discount of 30% , he will
5
7x
pays Rs. for n copies of books.
10
7x
B buys n books after paying Rs.
10
10n 4 x 4x
B buys books after paying Rs. (The amount which A spent to buy n
7x 5 5
copies of books )
According to question
8n
n 1 (The condition is given in the second part of the question)
7
n
or 1 or n 7
7
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PART ‘B’
21. Let A be a non-singular 3 3 matrix, the columns of which are denoted by the vectors
a , b and c , respectively. Similarly, u , v and w denote the vectors that form the
1
corresponding columns of AT . Which of the following is true?
(a) u a 0, u b 0, u c 1 (b) u a 0, u b 1, u c 0
(c) u a 1, u b 0, u c 0 (d) u a 0, u b 0, u c 0
Ans.: (c)
Solution: We can take any 3 3 non singular matrix in order to avoid long calculation.
1 0 0 1 0 0
0 2 0 0 1/ 2 0
1
Take A 0 0 3 AT 0 0 1/ 3
a
c u v
w
b
We see that
u .a 1.1 0.0 0.0 1
u .b 1.0 0.2 0.0 0
u .c 1.0 0.0 0.3 0
Q22.
Consider the real function f x 1/ x 2 4 . The Taylor expansion of f x about x 0
converges
(a) for all values of x (b) for all values of x except x 2
(c) in the region 2 x 2 (d) for x 2 and x 2
Ans.: (c)
1 1
Solution: f x
x 4
2
x2
4 1
4
1
1 x2
Thus the Taylor’s series of f x is times the binomial series of 1
4 4
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1 x 4 x 2 x 2 0
x 2
Now, the binomial series converges if
4
Hence x 2 0 2 x 2
1 1 1
1 2 3 0 4c 3b 2c 6 b 4 0
2 b 2c
2c 2b 2 0 b c 1
we do not need to perform further calculation.
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dy
Q24. Consider the differential equation ay e bt with the initial condition y 0 0 .
dt
Then the Laplace transform Y s of the solution y t is
1 1 1 e a eb
(a) (b) (c) (d)
s a s b b s a a s b ba
Ans.: (a)
dy
Solution: Given ay e bt
dt
Taking Laplace transform of both sides
We obtain
dy
dt
L aL y t L e bt sY s y 0 aY s
1
sb
Since, y 0 0 , we obtain
1 1
s a Y s Y s
sb s a s b
Q25. A light signal travels from a point A to a point B , both within a glass slab that is moving
with uniform velocity (in the same direction as the light) with speed 0.3c with respect to
an external observer. If the refractive index of the slab is 1.5 , then the observer will
measure the speed of the signal as
(a) 0.67 c (b) 0.81 c (c) 0.97 c (d) c
Ans.: (b)
Solution: v 0.3 c ,
S S
c
u x n 1.5
'
n
c 0 3c
0.3 c
u x' v n
c/n
ux '
u v 1 . c
c 0.3
1 x2
c n c2
u x 0.81 c .
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Q26. A monoatomic gas ‘of volume V is in equilibrium in a uniform vertical cylinder, the
lower end of which is closed by a rigid wall and the other by a frictionless piston. The
piston is pressed lightly and released. Assume that the gas is a poor conductor of heat and
the cylinder and piston are perfectly insulating. If the cross-sectional area of the cylinder
is A , the angular frequency of small oscillations of the piston about the point of
equilibrium, is
5
(a) 5 gA / 3V (b) 4 gA / 3V (c) gA / V (d) 7 gA / 5V
3
Ans.: (a)
Solution: PV P P V V
P V
PV P 1 V 1
P V
P V
PV PV 1 1
P V
P V P V
1 1 1 1 1 1
P V P V
V P P V
1 1 (i)
V P P V
P V
For small oscillation, also neglect
P V
F mg
From equilibrium P
A A
P V mg A dx
From (i), we get , P A mg. V , P A . A.
P V A V
mgdx mg
P A A F Adx
V V
gA 5 5 gA
and .
V 3 3 V
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Q27. A disc of mass m is free to rotate in a plane parallel to the xy plane with an angular
velocity ẑ about a massless rigid rod suspended from the roof of a stationary car (as
shown in the figure below). The rod is free to orient itself along any direction.
The car accelerates in the positive x -direction with an acceleration a 0 . Which of the
following statements is true for the coordinates
ẑ
of the centre of mass of the disc in the reference
ŷ
frame of the car?
(a) only the x and the z coordinates change x̂
(b) only the y and the z coordinates change
axˆ
(c) only the x and the y coordinates change
(d) all the three coordinates change
ẑ
Ans. : (d)
Q28. A cyclist, weighing a total of 80 kg with the bicycle, pedals at a speed of 10 m / s . She
stops pedalling at an instant which is taken to be t 0 . Due to the velocity dependent
10
frictional force, her velocity is found to vary as v t m.s , where t is
t
1
30
measured in seconds. When the velocity drops to 8 m / s , she starts pedalling again to
maintain a constant speed. The energy expended by her in 1 minute at this (new) speed, is
(a) 4 kJ (b) 8 kJ (c) 16 kJ (d) 32 kJ
Ans.: (b)
Solution: The acceleration of cyclist is
d 300 300
a t
dt t 30 t 30 2
Since frictional force is the only force acting on the cyclist, this net force is equal to
functional force. We can write
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80 300 4
2
F t v t
2
300 t 30 15
4 2
When the cyclist moves at a constant speed, the frictional force is F t 8 .
15
The displacement during this interval is 8 60 . Thus the work done by frictional force
4
64 8 60 is 8.192 KJ . Hence in order to maintain constant speed the cyclist
15
must supply an energy equal to 8.192 KJ .
Q29. The number of linearly independent power series solutions, around x 0 , of the second
d 2 y dy
order linear differential equation x xy 0 , is
dx 2 dx
(a) 0 (this equation does not have a power series solution)
(b) 1
(c) 2
(d) 3
Ans. : (b)
Q30. Let x denote the position operator and p the canonically conjugate momentum operator
of a particle. The commutator
1 2 2 1 2
2m p x , m p x
2
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Q31. Two point charges 3Q and Q are placed at 0, 0, d and 0, 0, 2d respectively,
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Q33. An electromagnetic wave is travelling in free space (of permittivity 0 ) with electric field
ˆ cos q x ct
E kE 0
The average power (per unit area) crossing planes parallel to 4 x 3 y 0 will be
4 1 16
(a) 0 cE02 (b) 0 cE02 (c) 0 cE02 (d) 0 cE02
5 2 25
Ans.: (c)
x y
Solution: 4 x 3 y 0 0
3 4 y
n̂
E
B 0 cos qx qct yˆ 4
c
1 1 B 90
S
0
EB
E0
0 0 c
E cos x S
2
ˆ
E02
2 0 c
xˆ
x
K 3
E2 E2 2
I S .nˆ 0 cos 90 0 sin c 0 E02
2 0 c 2 0 c 5 E
z
4 4
tan sin
3 5
1
I 0.4c 0 E02 c 0 E02
2
Q34. A plane electromagnetic wave from within a dielectric medium (with 4 0 and 0 )
is incident on its boundary with air, at z 0 . The magnetic field in the medium is
H ˆjH 0 cos t kx k 3 z , where and k are positive constants.
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k 1
sin T 2sin I tan I x I 300
kz 3
sin T 2 sin 30 1 T 90
0 0
distance from the origin to the mean distance from the origin, is
[You may use dx x n e x n ! ]
0
1 1 3 2
(a) (b) (c) (d)
3 2 2 3
Ans.: (d)
r
1
Solution: r , , e 2a
8 a 3
3
r r * r 2 dr sin d d 2a 3a
2
one can compare the wave function at hydrogen atom with Bohr radius a0 2a
most probable distance,
d 2 r / a
r e 0
dr
rP 2a
rp 2a 2
r 3a 3
Q36. The state vector of a one-dimensional simple harmonic oscillator of angular frequency ,
1
at time t 0 , is given by 0 0 2 , where 0 and 2 are the normalized
2
ground state and the second excited state, respectively. The minimum time t after which
the state vector t is orthogonal to 0 , is
2 4
(a) (b) (c) (d)
2
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Ans.: (a)
1
Solution: 0 0 2
2
t 5 t
5 1
E2 t 0 e 2
2 e 2
2 2
E0 0 t 0 t cos 1 1
2 E2 E0
t cos 1 1 .
5 1 2 / 2 2
2 2
Q37. The normalized wavefunction in the momentum space of a particle in one dimension is
p , where and are real constants. The uncertainty x in measuring
p 2
2
its position is
(a) (b) (c) (d)
2
3
2
Ans.: (c)
Solution: p
p 2
2
2
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Q38. The full scale voltage of an n -bit Digital-to-Analog Convener is V . The resolution that
can be achieved in it is
V V V V
(a) (b) (c) (d)
2 1
n
2 1
n
22 n n
Ans. : (a)
Q39. The spring constant k of a spring of mass ms is determined experimentally by loading
the spring with mass M and recording the time period T , for a single oscillation. If the
experiment is carried out for different masses, then the graph that correctly represents the
result is
(a) (b)
T2 T2
0, 0 M 0, 0 M
(c) T2 (d) T2
0, 0 M 0, 0 M
Ans.: (a)
Solution: The Langragian of system.
1 ms 2 1 1 d L L
L x Mx 2 kx 2 , 0
2 3 2 2 dt x x
d L m
0 s M
x kx
dt x 3
ms ms
M M 3
T 2 3 T 2 4
k k
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Q40. A Zener diode with an operating voltage of 10 V at 250 C has a positive temperature
coefficient of 0.07 % per 0 C of the operating voltage. The operating voltage of this
E dE L 4
2 2
L
g E dE 2 .2 . . . E.dE.
2 v v 2 v 2
at T 0 K ,
L2 4 2F
EF
N g E dE .
0
4 2 v 2 2
L2
N 2F
2 v
2 2
N 2F
n
L2 2 v 2 2
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Q42. The relation between the internal energy U , entropy S , temperature T , pressure p ,
volume V , chemical potential and number of particles N of a thermodynamic system
is dU TdS pdV dN . That U is an exact differential implies that
p T U U
(a) (b) p S
S V , N V S ,N T S ,N V S ,
U 1 U p T
(c) p (d)
T S ,N T V S , S V , N V S ,N
Ans.: (a)
Solution: df Adx Bdy Cdz
If f is perfect differential then, A T , B p, C
x S y V, z N
A B
y x , z x y , z
A C B C
,
z x , y x y , z z x , y y x , z
T P
V S , N S V , N
Q43. The number of microstates of a gas of N particles in a volume V and of internal energy
U , is given by
3N / 2
N aU
U , V , N V Nb
N
(where a and b are positive constants). Its pressure P , volume V and temperature T ,
are related by
aN aN
(a) P V Nb Nk BT (b) P 2 V Nb Nk BT
V V
(c) PV Nk BT (d) P V Nb Nk BT
Ans.: (d)
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3N
aU
Solution: u , v, N V Nb
N 2
N
3 aU
S k ln Nk ln V Nb ln
2 N
1 P
dS dU dV
T T
S 1 3
U NkT
U V T 2
S P P Nk
V U T T V Nb
P V Nb NkT
Q44. Consider a system of identical atoms in equilibrium with blackbody radiation in a cavity
at temperature T . The equilibrium probabilities for each atom being in the ground state
0 and an excited state 1 are P0 and P1 respectively. Let n be the average number of
photons in a mode in the cavity that causes transition between the two states. Let W01
and W10 denote, respectively, the squares of the matrix elements corresponding to the
equilibrium?
(a) P0 nW01 PW
1 1 0 (b) P0 nW01 PnW
1 1 0
(c) P0 nW01 PW
1 10 PnW
1 10 (d) P0 nW01 PW
1 10 PnW
1 10
Ans.: (d)
Solution: In equilibrium condition the number of upward transitions must be equal to the number
of downward transition.
Rate of upward transition P0 nW01
Rate of downward transition = Rate of spontaneous transitions + Rate of stimulated
transition PW
1 10 PnW
1 10 1
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P0 nW01 PW1 10 PnW
1 10
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PART ‘C’
Q46. Consider a set of particles which interact by a pair potential V ar 6 where r is the inter-
particle separation and a 0 is a constant. If a system of such particles has reached virial
equilibrium, the ratio of the kinetic to the total energy of the system is
1 1 3 2
(a) (b) (c) (d)
2 3 4 3
Ans.: (c)
n
Solution: V r ar n T V
2
V r a r 6 and T 3 V
1 4 T 3
Then, E T V T T E T
3 3 E 4
Q47. In an inertial frame S , the magnetic vector potential in a region of space is given by
A az iˆ (where a is a constant) and the scalar potential is zero. The electric and
magnetic fields seen by an inertial observer moving with a velocity viˆ with respect to S ,
1
are, respectively [In the following ]
v2
1 2
c
v v
Bx Bx , By By 2 Ez and Bz Bz 2 E y
c c
A
E V 0, B A a ˆj
t
E x 0, E y 0 v 0 0, Ez 0 va va
(replace v by v ) E v azˆ
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v v
Bx 0, By a 2 0 a, Bz 0 2 0 0
c c
B a ˆj
Q48. In the rest frame S1 of a point particle with electric charge q1 another point particle with
distance l . The magnitude of the electromagnetic force felt by q1 due to q2 when the
1
distance between them is minimum, is [In the following ]
v2
1 2
c
1 q1q2 1 q1q2
(a) (b)
4 0 l 2 4 0 l 2
1 q1q2 v 2 1 q1q2 v 2
(c) 1 (d) 1
4 0 l 2 c 2 4 0 l 2 c 2
Ans.: (b)
1 q1q2
Solution: Charge of q2 seen by rest frame of q1 q2 ; F
4 0 l 2
Q49. A circular current carrying loop of radius a carries a steady current. A constant electric
charge is kept at the centre of the loop. The electric and magnetic fields, E and B
respectively, at a distance d vertically above the centre of the loop satisfy
(a) E B (b) E 0 (c) E B 0
(d) E B 0
Ans.: (c)
Solution: E B 0 . E B 0
Q50. A phase shift of 300 is observed when a beam of particles of energy 0.1 MeV is scattered
by a target. When the beam energy is changed, the observed phase shift is 600 . Assuming
that only s -wave scattering is relevant and that the cross-section does not change with
energy, the beam energy is
(a) 0.4 MeV (b) 0.3 MeV (c) 0.2 MeV (d) 0.15 MeV
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Ans. : (b)
4
Solution: 2
2l 1 sin 2 l
k l 0
2mE
only s -wave scattering is relevant l 0 k
2
4 4 2
2 sin 0
2
sin 2 0
k 2mE
always be
(a) positive definite (b) real symmetric (c) hermitian (d) anti-hermitian
Ans.: (d)
Solution: Since, U e im
U U 1 X d U
or, Xd im d
here, m is one of the Pauli spin matrices, since Pauli matrices are hermitian,
taken complex conjugate, so matrix should anti-hermitian.
Hence correct option is (d)
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dy x
Q52. The differential equation x 2 , with the initial condition y 0 0 , is solved
dx
using Euler’s method. If yE x is the exact solution and y N x the numerical solution
Euler’s method, yi yi 1 hf xi 1 , yi 1
y1 0 , y2 h3 y3 5 h3
yn
n 1 n 2n 1 h3
6
Since, 0,5,14,30,... different from square terms
At, x0 0 x1 x0 h h x2 x0 2h 2h x3 x0 3h 3h
xn 1 x0 n 1 h n 1 h . Now, xn nh
f x0 , y0 0 , f x1 , y1 h 2 , f x2 , y2 4 h 2
f xn 1 , yn 1 n 1 h 2
2
n 1 n 2n 1 h3 n3h3
y N yE 6 3
yE n3 h3
3
yN yE 1
By solving,
yE n
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Q53. The interval 0,1 is divided into n parts of equal length to calculate the integral
1
e
i 2 x
dx using the trapezoidal rule. The minimum value of n for which the result is
0
exact, is
(a) 2 (b) 3 (c) 4 (d)
Ans.: (a)
1
Solution: ei 2 nx dx 0 , exact value
0
1
Now, nh 1 0, h
n
1
y f x ei 2 n x , Let n 2 , then x0 0, y0 1 , x1 , y1 1 and x2 1, y2 1
2
h 1
I y0 2 y1 .... yn 1 yn 1 2 1 1
2 4
I 0. So, n 2
Q54. The generating function G t , x for the Legendre polynomials Pn t is
1
G t, x x n Pn t , for x 1
1 2 xt x 2 n 0
x
If the function f x is defined by the integral equation f x dx xG 1, x , it can be
0
expressed as
1
(a)
n ,m 0
x n m Pn 1 Pm
2
(b)
n ,m 0
x n m Pn 1 Pm 1
(c)
n ,m 0
x n m Pn 1 Pm 1 (d)
n ,m 0
x n m Pn 0 Pm 1
Ans. : (b)
1
Solution: G t , x x n Pn t for x 1
1 2 xt x 2 n 0
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1
G 1, x x n Pn 1
1 2x x 2
n0
1 1
x n Pn 1 Since x 1
1 x
2
n 0 1 x
x
1
f x dx
1 x 0
Now, x
are constants. Which of the following curves best describes the trajectories of this system
in phase space?
p p
(a) (b)
x x
p
p
(c) (d)
x x
Ans.: (c)
Solution: V x k 2 x 4 2 x 2
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For equation point
V 2 V x
0 4k 2 x3 2 2 x 0 , x 0 or x 2 2
x 2k
d 2V x
Now, 12k 2 x 2 2 2 At, x 0
dx 2
Px
d 2V
2 2 , x 0 is minimum.
dx 2
x
dV2
2
2
And, 2
12k 2 2 2 2 4 2 , at x 2 2
dx 2k 2k
2
Hence, x is maxima.
2k 2
Q56. Let x, p be the generalized coordinate and momentum of a Hamiltonian system. If new
and are constants, then the conditions for it to be a canonical transformation, are
1 1 1 1
(a) 1 and 1 (b) 1 and 1
2 2 2 2
1 1 1 1
(c) 1 and 1 (d) 1 and 1
2 2 2 2
Ans.: (c)
Solution: X x sinh p
P x cosh p
For canonical transformation
X P X P
. . 1
x p p X
1 0 (i)
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sin h p v cos h p cos h p sin h p
2 2 2 2
energy E 0 , is
1 1 1 1
(a) (b) (c) cosh 2 (d)
1 2 cosh 2 2 cosh 2 2 cosh 2
Ans.: (a)
kT
e e 1 2 cosh
Solution: Partition function is z 1 e e kT z 1 2
kT kT
2 kT
Probability that system has energy, E 0
1
P E 0
1 2 cosh
kT
put 2kT
1
P E 0
1 2 cosh 2
Q58. Two non-degenerate energy levels with energies 0 and are occupied by N non-
interacting particles at a temperature T . Using classical statistics, the average internal
energy of the system is
N N 3
(a) (b) (c) N e/ kBT (d) Nk BT
1 e/ kBT
1 e/ kBT 2
Ans.: (a)
Solution: For one particle
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exp
Quantum mechanical energy is, U kT
1 e kT
N exp
kT U N
For N particle, U
1 exp 1 exp
kT kT
Q59. In the circuit below, D1 and D2 are two silicon diodes with the same characteristics. If
the forward voltage drop of a silicon diode is 0.7 V then the value of the current I1 I D1
I1
is
1k I D1 ID2
10V D1 D2 Vo
1 1 1
Q60. The Hamiltonian of a two-level quantum system is H possible initial
2 1 1
state in which the probability of the system being in that quantum state does not change
with time, is
cos 4 cos 8 cos 2 cos 6
(a) (b) (c) (d)
sin sin sin sin
4 8 2 6
Ans. : (b)
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Q61. Consider a one-dimensional infinite square well
0 for 0 x a,
V x
otherwise
If a perturbation
V for 0 x a / 3,
V x 0
0 otherwise
is applied, then the correction to the energy of the first excited state, to first order in V ,
is nearest to
(a) V0 (b) 0.16 V0 (c) 0.2 V0 (d) 0.33 V0
Ans. (d)
a/3
Solution: V V dx
*
x 2 2
0
2 2 2 x 1 4 x
a/3 a/3
2
V V0
0
a
sin
a
dx V0
a
0
2
1 cos
a
dx
4
sin 1
2 a 3 3
V0
4 V0 0.33 V0
a 6 3 4
a
Q62. The energy eigenvalues En of a quantum system in the potential V cx 6 (where c 0 is
31
1/ 2 1/ 6
E n E 6
n , E n3/ 2
Therefore, correct option is (b)
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Q63. Consider a quantum system of non-interacting bosons in contact with a particle bath. The
probability of finding no particle in a given single particle quantum state is 106 . The
average number of particles in that state is of the order of
(a) 103 (b) 106 (c) 109 (d) 1012
Ans. : (b)
Q64. The sensitivity of a hot cathode pressure gauge is 10 mbar 1 . If the ratio between the
numbers of the impinging charged particles to emitted electrons is 1:10 , then the
pressure
(a) 10 mbar (b) 101 mbar (c) 102 mbar (d) 102 mbar
Ans.: (c)
I 1
Solution: Pressure, P
I S
I
where ratio between the number of the impinging charge particles to emitted
I
electrons
S Sensitivity of Gauge.
1 1
P 1
102 m bar
10 10 m bar
2 M b
Q65. Two physical quantities T and M are related by the equation T , where a
a 2
and b are constant parameters. The variation of T as a function of M was recorded in
an experiment to determine the value of a graphically. Let m be the slope of the straight
line when T 2 is plotted vs M , and m be the uncertainty in determining it. The
uncertainty in determining a is
a m m b m 2 m
(a) (b) a (c) (d)
2 m m 2a m a m
Ans.: (a)
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2 M b 2 2
2 b
2
Solution: T T 2 2 M 2 mM C
a 2 a a
2 2 2
where slope is m 2 a 1/ 2
a m
a 1
Now
m 2m
3/ 2
a 2 2 m
2 2
a
2
2
2
m m
m 2 m3 2m m
m a m a m
a a
2 m m 2 2 m 2 m
Thus correct option is (a)
Q66. The circuit below comprises of D -flip flops. The output is taken from Q3 , Q2,Q1 and Q0 as
LSB Q Q1 Q2
MSB Q
0 3
D Q D Q D Q D Q
CLK
RST
the binary number given by the string Q3 , Q2,Q1Q0 changes for every clock pulse that is
applied to the CLK input. If the output is initialized at 0000 , the the corresponding
sequence of decimal numbers that repeats itself, is
(a) 3, 2,1, 0 (b) 1,3, 7,14,12,8
(c) 1,3, 7,15,12,14, 0 (d) 1,3, 7,15,14,12,8, 0
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Ans.: (d)
Solution: Clock Q3 Q2 Q1 Q0
0 0 0 0
1 0 0 0 1 1
2 0 0 1 1 3
3 0 1 1 1 7
4 1 1 1 1 15
5 1 1 1 0 14
6 1 1 0 0 12
7 1 0 0 0 8
8 0 0 0 0 0
Q67. The spin-parity assignments for the ground and first excited states of the isotope 57
28 Ni , in
3
So, j , l 1
2
3
Spin parity for ground state of 28 Ni 57
2
For first excited state,
1s1/2 21 p3/4 21 p1/2 21d5/6 2 2s1/2 21d3/4 21 f 78/ 2 2 p3/1 2 1 f 5/ 2
5 5
P , l 3 spin parity
2 2
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238
Q68. The first excited state of the rotational spectrum of the nucleus 92 U has an energy
45 keV above the ground state. The energy of the second excited state (in keV) is
(a) 150 (b) 120 (c) 90 (d) 60
Solution: As per the shell model (Collective Model)
Rotational Energies,
2
Er J J 1 , I is moment of inertia where only even value of J are allowed
2I
i.e., J 0 , 2 , 4 , 6 ,........
Now, for ground state J 0 , E 0 keV
(c) K 0 p n (d) p n
Ans.: (a)
Solution: (1) K0 0 K0
Charge 0 0 0 1 1 Conserved
Spin 0 0 0 0 0 Conserved
1 1
I 1 1 1 Not conserved
2 2
1 1
I3 0 1 1 I3 1
2 2
S 1 0 1 0 0 S 1
This interaction is not allowed by strong interaction but allowed by weak interaction.
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Q70. A crystal of MnO has NaCl structure. It has a paramagnetic to anti- ferromagnetic
transition at 120 K . Below 120 K , the spins within a single 111 planes are parallel but
the spins in adjacent 111 planes are antiparallel. If neutron scattering is used to
determine the lattice constants, respectively, d and d , below and above the transition
temperature of MnO then
d d
(a) d (b) d (c) d 2d (d) d 2d
2 2
Ans.: (c)
Solution: At 80 K there are extra reflection not present at 293 K .
The reflection at 80 K may be classified in term of a cubic unit cell at lattice constant
0
8.85 A .
0
At 293 K the reflection corresponds to an FCC unit cell of lattice constant 4.43 A
d a / 3 4.43 1
d 2d
d a / 3 8.85 2
Q71. A metallic nanowire of length l is approximated as a one-dimensional lattice of N
atoms with lattice spacing a . If the dispersion of electrons in the lattice is given
as E k E0 2t cos ka , where E0 and t are constants, then the dnsity of states inside
E E0
2
t2
2t 1
3
(a) N (b)
E E0
E E0 N
(c) N 3 (d)
t2 2t E E0
2 2
Ans. (d)
Solution: E k E0 2t cos ka
dE
dE 2ta sin ka dk dk
2 ta sin ka
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2L 2L dE
g E dE dk
2 ta 1 cos 2 ka
L dE 2L dE
ta E E
2 a 2t E E0
2 2
1 0
2t
Density of state is,
g E dE 2N
E
dE 2t E E0
2 2
N
E
2t E E0
2 2
2 k l 2 w 2 2 k
(a) (b) (c) (d)
q2 w kq 2 l kq 2 q2
Ans.: (c)
A VH A
Solution: xy R A lw
l I l
IB
where, VH
nqw
B A B B 2
xy 2
nqw l nq k q B q kq
2
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Q73. The Zeeman shift of the energy of a state with quantum numbers L, S , J and mJ is
mJ B B
Hz
J J 1
L.J g s S .J
where B is the applied magnetic field, g s is the g -factor for the spin and
B
1.4 MHz G 1 , where h is the Planck constant. The approximate frequency shift of
h
the S 0, L 1 and mJ 1 state, at a magnetic field of 1G , is
(a) 10 MHz (b) 1.4 MHz (c) 5 MHz (d) 2.8 MHz
Ans.: (b)
eB e B B J 2
Solution: v . B E2
4 m 2m h h J 1
1.4 MH z G 1
1G 1.4 MHz E1
J
Q74. The separations between the adjacent levels of a normal multiplet are found to be
22 cm 1 and 33 cm 1 . Assume that the multiplet is described well by the L S coupling
Ans.: (d)
Solution: E1 E J 1 E J A J 1 22 cm 1
E2 E J 2 E J 1 A J 2 33 cm 1
A J 1 22
J 1
A J 2 33
Thus, the J values for all those three levels are 1, 2,3 . The, corresponding terms are
3
D1,2,3 .
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Q75. If the fine structure splitting between the 2 P3/ 2 and 2 2 P3/ 2 levels in the hydrogen atom
2
E Li zLi 4 34
81
E H zH 4 14
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MATHEMATICAL PHYSICS SOLUTIONS
NET/JRF (JUNE-2011)
Q1. The value of the integral dz z 2 e z , where C is an open contour in the complex z -plane as
C
lm z
shown in the figure below, is:
5 5
(a) e (b) e 0,1
e e
5 5 C
(c) e (d) e
e e
Re z
Ans: (c) 1,0 1,0
Solution: If we complete the contour, then by Cauchy integral theorem
1 1
e dzz e 0 dzz e dzz 2 e z z 2 e z 2 ze z 2e z 1
1 5
dzz e
2 z 2 z 2 z
1 C C 1
e
Q2. Which of the following matrices is an element of the group SU 2 ?
1 i 1
1 1
(a) (b) 3 3
0 1 1 1 i
3 3
1 3
2 i i
(c) (d) 2 2
3 1 i 3 1
2 2
Ans: (b)
Solution: SU 2 is a group defined as following: SU 2
2 2
: , C ; 1
1 i 1 1 i 1
clearly (b) hold the property of SU 2 . , and , .
3 3 3 3
Note: SU 2 has wide applications in electroweak interaction covered in standard model
of particle physics.
Q3. Let a and b be two distinct three dimensional vectors. Then the component of b that is
perpendicular to a is given by
(a)
a ba
(b)
b a b
(c)
a b b (d)
b a a
a2 b2 b2 a2
Ans: (a)
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Solution: a b ab sin nˆ where n̂ is perpendicular to plane containing
a and b and pointing upwards.
b b sin kˆ
a a b ab sin a nˆ a 2 b sin kˆ
ˆ
b sin k
a a b ˆ
b sin k
a ba
.
a
a2 a2
Q4. Let p n x (where n 0,1, 2, ...... ) be a polynomial of degree n with real coefficients,
4
defined in the interval 2 n 4 . If pn x pm x dx nm , then
2
1 3 1
(a) p 0 x and p1 x 3 x (b) p0 x and p1 x 3 3 x
2 2 2
1 3 1 3
(c) p0 x and p1 x 3 x (d) p 0 x and p1 x 3 x
2 2 2 2
Ans: (d)
Solution: For n not equal to m kroneker delta become zero. One positive and one negative term
1
can make integral zero. So answer may be (c) or (d). Now take n m 0 so p0 x
2
and then integrate. (d) is correct option because it satisfies the equation Check by
integration and by orthogonal property of Legendre polynomial also.
Q5. Which of the following is an analytic function of the complex variable z x iy in the
domain z 2 ?
(a) 3 x iy (b) 1 x iy 7 x iy
7 4 3
(c) 1 x iy 7 x iy (d) x iy 1
4 3 1/ 2
Ans: (b)
Solution: Put z x iy . If z x iy appears in any of the expressions then that expression is
1
non-analytic. For option (d) we have a branch point singularity as the power is which
2
is fractional. Hence only option (b) is analytic.
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1 1 1
Q6. Consider the matrix M 1 1 1
1 1 1
A. The eigenvalues of M are
(a) 0, 1, 2 (b) 0, 0, 3 (c) 1, 1, 1 (d) – 1, 1, 3
Ans: (b)
1 1 1
Solution: For eigen values 1 1 1 0
1 1 1
1 1 2 1 1 1 11 1 0
1 1 2 2 1 0 2 2 3 22 2 0
3 32 0 2 3 0 0, 0, 3
For any n n matrix having all elements unity eigenvalues are 0, 0, 0,..., n .
B. The exponential of M simplifies to (I is the 3 3 identity matrix)
e3 1 M2
(a) e I
M
M (b) e I M
M
3 2!
2 1 1 x1 0
For 3 , 1 2 1 x 2 0
1 1 2 x3 0
2 x1 x2 x 3 0 , x1 2 x 2 x3 0 , x1 x 2 2 x3 0
3x 2 3x3 0 or x2 x3 x1 x 2 x3 k .
1
1
Eigen vector is 1 , where k 1 .
3
1
For 0 ,
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1 1 1 x1 0
1 1 1 x 0 x x x 0
2 1 2 3
k1 1
1
Let x1 k1 , x2 k2 and x3 k1 k2 . Eigen vector is k2 1 where k1 k 2 1 .
k 1 k 2 2
1
1
1
Let x1 k1 , x 2 k 2 and x3 k1 k 2 . Other Eigen vector 0 where k1 1, k 2 1 .
2
1
0 1 1 1 2 1
S 1 0 1 S 2 1 1 D S 1 MS , M SDS 1 .
1
1 1 1 1 1 1
1 0 0
eM Se D S 1 e D 0 1 0 e M 1
e3 1 M
3
0 0 e 3
NET/JRF (DEC-2011)
Q7. An unbiased dice is thrown three times successively. The probability that the numbers of
dots on the uppermost surface add up to 16 is
1 1 1 1
(a) (b) (c) (d)
16 36 108 216
Ans: (b)
Solution: We can get sum of dice as 16 in total six ways i.e. three ways (6, 5, 5) and three ways
(6, 6, 4).
Total number of ways for 3 dice having six faces 6 6 6
6 1
6 6 6 36
Q8. The generating function F x, t Pn x t n for the Legendre polynomials Pn x
n 0
is F x, t 1 2 xt t 2
1
2
. The value of P3 1 is
(a) 5 / 2 (b) 3 / 2 (c) 1 (d) 1
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Ans: (d)
Solution: P3
1
2
1
2
1
5 x 3 3x P3 1 5 1 3 1 5 3 1
3
2
Q9. The equation of the plane that is tangent to the surface xyz 8 at the point 1, 2, 4 is
(a) x 2 y 4 z 12 (b) 4 x 2 y z 12
(c) x 4 y 2 0 (d) x y z 7
Ans: (b)
Solution: To get a normal at the surface, lets take the gradient
Q11. Let x1 t and x2 t be two linearly independent solutions of the differential equation
d 2x dx t dx t
2 f t x 0 and let wt x1 t 2 x2 t 1 . If w0 1, then w1 is
dx
2
dt dt dt dt
given by
(a) 1 (b) e 2 (c) 1 / e (d) 1 / e 2
Ans: (d)
Solution: W t is Wronskian of D.E.
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1 for 2n x 2n 1
Q12. The graph of the function f x
0 for 2n 1 x 2n 2
~
where n 0,1, 2,...... is shown below. Its Laplace transform f s is
f x
1 es 1 es
(a) (b)
s s 1
1 1
(c) (d) x
s 1 e s s 1 e s 0 1 2 3 4 5
Ans: (c)
1 2 3
Solution: L f x e sx
f x dx e sx
1dx e sx
0dx e sx 1dx ......
0 0 1 2
1 3
e sx e sx
0 ......
1 s
e 1
1 3 s
e e 2 s ......
s 0 s 2 s s
1
s
1
1 e s e 2 s e 3 s ........ 1 e s e 2 s e 3s ....
s
a 1 1
Since S where r e s and a 1 S .
1 r s 1 e s
Q13. The first few terms in the Taylor series expansion of the function f x sin x around
x are:
4
1 1 1
2 3
(a)
1 x x x .....
2 4 2! 4 3! 4
1
2 3
1 1
(b) 1
x x x .....
2 4 2! 4 3! 4
1
3
(c) x x .....
4 3! 4
1 x 2 x3
(d) 1 x .....
2 2! 3!
Ans: (b)
Solution: f x sin x
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1 1 1
f , f cos , f sin
4 2 4 4 2 4 4 2
1
2 3
1 1
So Taylor’s series is given by 1 x x x .....
2 4 2! 4 3! 4
NET/JRF (JUNE-2012)
Q14. A vector perpendicular to any vector that lies on the plane defined by x y z 5 , is
1 2 3
Q15. The eigen values of the matrix A 2 4 6 are
3 6 9
(a) 1, 4, 9 (b) 0, 7, 7 (c) 0,1,13 (d) 0, 0,14
Ans: (d)
1 2 3
Solution: For eigenvalues A I 0 2 4 6 0
3 6 9
1 z z 2 ....1 z z z ....
2 3
1 1
z 1 z 1 z ....
2 3
(a) (b)
2 2 4 8 1 z
1 1 1 2 4
(d) 2 z 1 5 z 1 7z 1 ....
2 3
(c)
z2 1 z z 2 .... 1 z z 2 ....
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Ans: (b)
1 1 1 1 1 1
1 1 z
1
Solution:
z 1z 2 z 2 z 1 1 z z 1 1 1 z
1
1 1 z
1 2 1 z 2 1 2 3 1 z 3 ...
1 z 2! 3!
1
1 z
z 1 z 1 z ....
2 3
1 2
Q17. Let u x, y x x y 2 be the real part of analytic function f z of the complex
2
variable z x i y . The imaginary part of f z is
Solution: u x, y x
1 2
2
x y 2 , v x, y ?
u v u v
Check and .
x y y x
u v v
, 1 x , v y xy f x
x y y
u v v
y, v yx f y
y x x
y xy f x yx f y
If f x 0, f y y
v xy y
Q18. Let y x be a continuous real function in the range 0 and 2 , satisfying the
d2y dy
inhomogeneous differential equation: sin x 2
cos x x
dx dx 2
The value of dyldx at the point x / 2
(a) is continuous (b) has a discontinuity of 3
(c) has a discontinuity of 1/3 (d) has a discontinuity of 1
Ans: (d)
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2
d y dy
Solution: After dividing by sin x , 2
cot x cosec x x
dx dx 2
dy dy
Integrating both sides, cot x dx cosec x x dx
dx dx 2
dy
cot x y cosec 2 x ydx 1
dx
dy cos x
y y cosec 2 xdx 1 , at x ; sin x 0 . So this is point of discontinuity.
dx sin x
Q19. A ball is picked at random from one of two boxes that contain 2 black and 3 white and 3
black and 4white balls respectively. What is the probability that it is white?
(a) 34 / 70 (b) 41 / 70 (c) 36 / 70 (d) 29 / 70
Ans: (b)
Solution: Probability of picking white ball
2 B 3W 3B 4W
3 4
From box I and from box II
5 7
1 3 4 41
Probability of picking a white ball from either of the two boxes is
2 5 7 70
Q20. The eigenvalues of the antisymmetric matrix,
0 n3 n2
A n3 0 n1
n2 0
n1
0 n3 n2
A I 0, n3 0 n1 0
n2 n1 0
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1 0 2 n12 n22 n32 3 n12 n22 n32
so, 1 0 , 2 i , 3 i
Ans: (b)
Q22. A bag contains many balls, each with a number painted on it. There are exactly n balls
which have the number n (namely one ball with 1, two balls with 2, and so on until N on
them). An experiment consists of choosing a ball at random, noting the number on it and
returning it to the bag. If the experiment is repeated a large number of times, the average
value the number will tend to
2N 1 N N 1 N N 1
(a) (b) (c) (d)
3 2 2 2
Ans: (a)
N N 1
Solution: Total number of balls 1 2 3 4 ..... N
2
k
The probability for choosing a k th ball at random
N N 1
2
2 k 2 2 N N 12 N 1
Average of it is given by k k P
N N 1 N N 1 6
2N 1 N N 12 N 1
where k 2 .
3 6
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Q23. Consider a sinusoidal waveform of amplitude 1V and frequency f 0 . Starting from an
1
arbitrary initial time, the waveform is sampled at intervals of . If the corresponding
2 f0
NET/JRF (DEC-2012)
a b c
Q24. The unit normal vector of the point , , on the surface of the ellipsoid
3 3 3
x2 y 2 z 2
1 is
a2 b2 c2
bciˆ caˆj abkˆ aiˆ bˆj ckˆ
(a) (b)
a2 b2 c2 a2 b2 c2
Unit normal vector is .
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x 2
y 2
z2 2 xiˆ 2 yˆj 2 zkˆ
So, i ˆj kˆ 2 2 2 1 2 2 2
x y z a b c a b c
2 ˆ 2 ˆ 2 ˆ
a bc
i j k
, ,
3 3 3
a 3 b 3 c 3
4 4 4 2 b 2c 2 a 2c 2 a 2c 2
3a 2 3b 2 3c 2 3 a 2b 2 c 2
2 ˆ 2 ˆ 2 ˆ
i j k
a 3 b 3 c 3 bciˆ caˆj abkˆ
a b c
2 b 2c 2 c 2 a 2 a 2b 2 b 2c 2 c 2 a 2 a 2b 2
, ,
3 3 3 3 abc
0.00
(c) e x cos x
0.25
0.50
1
(d) cos x 0.75
x 1.00
0 1 2 3 4 5 6 7 8 9 10
Ans: (a) x
Q27. In a series of five Cricket matches, one of the captains calls “Heads” every time when the
toss is taken. The probability that he will win 3 times and lose 2 times is
(a) 1 / 8 (b) 5 / 8 (c) 3 / 16 (d) 5 / 16
Ans: (d)
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3 53 2
1 1 5! 1 1 5!
Solution: P 1 .
2 2 3!5 3! 8 2 3!5 3!
1 5 4 3! 20 5 5
32 3! 2! 32 2 8 2 16
Q28. The Taylor expansion of the function ln cosh x , where x is real, about the point x 0
starts with the following terms:
1 2 1 4 1 2 1 4
(a) x x .... (b) x x ....
2 12 2 12
1 2 1 4 1 2 1 4
(c) x x .... (d) x x ....
2 6 2 6
Ans: (b)
e x ex
Solution: cosh x .Tailor’s series expansion of f x about x a
2
f a f '' a f '''
f x f a x a x a x a ... . Here a 0 .
2 3
1 ! 2 ! 3 !
e x ex 1 e x e x e x e x
f x log 0 , f x x 0 x x x tanh x 0
2 x 0 e e 2 e e x
2
f ' ' x
e x
ex e x ex e x ex e x ex e x
ex e
2 x
ex
2
1 tanh 2 x
e x
e x 2
e x
e
x 2
1 2 1 4
f x x x .......
2 12
z 3 dz
Q29. The value of the integral , where C is a closed contour defined by the
C z 2
5z 6
equation 2 z 5 0, traversed in the anti-clockwise direction, is
z3 8 z 3 dz
Residue z 2 8 2 2 i 8 16 i
z 3z 2 z 2 2 3 c z 5z 6
NET/JRF (JUNE-2013)
tn
H n x
2
Q30. Given that e t 2tx
n 0 n!
the value of H 4 0 is
(a) 12 (b) 6 (c) 24 (d) – 6
Ans: (a)
tn tn t4 t6
H n x e t 2tx H n 0 e t 1 t 2
2 2
Solution:
n 0 n! n 0 n! 2! 3!
H 4 0 4 t 4 4!
t H 4 0 12 .
4! 2! 2!
Q31. A unit vector n̂ on the xy -plane is at an angle of 120 o with respect to iˆ . The angle
between the vectors u a iˆ b nˆ and v anˆ b iˆ will be 60 o if
u v aiˆ bnˆ anˆ biˆ u v cos 60 a 2 iˆ nˆ ab ba b 2 nˆ.iˆ
a 2
b 2 2ab cos120 cos 60 a
2
2
cos120 2ab b 2 cos120
2 2 1
2
1 2
2
2 1
1
a b 2ab cos 60 a b 2ab a 2 b 2 a 2 b 2 2ab
ab
2 2 2
5ab a
a2 b2 b .
2 2
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Q32. With z x iy, which of the following functions f x, y is NOT a (complex) analytic
function of z ?
(a) f x, y x iy 8 4 x 2 y 2 2ixy
3 7
(b) f x, y x iy 1 x iy
7 3
(c) f x, y x 2 y 2 2ixy 3
5
(d) f x, y 1 x iy 2 x iy
4 6
Ans: (d)
Solution: f x, y 1 x iy 2 x iy 1 x iy 2 x iy
4 6 4 6
Due to present of z x iy
Q33. The solution of the partial differential equation
2 2
u x , t u x, t 0
t 2 x 2
satisfying the boundary conditions u 0, t 0 u L, t and initial conditions
u x,0 sin x / L and u x, t t 0 sin 2x / L is
t
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n x n x
L L
2 2
with An f x sin dx, Bn g x sin dx
L0 L an 0 L
2u 2u x 2 x
Comparing a 2 2 , We have a 1 and f x sin , g x sin ,
t 2
x L L
2 x
1 cos
x n x x L dx 2 L 1 (let n 1 )
L L L
2 2 2
An sin sin dx sin 2 dx
L0 L L L0 L L 0 2 L 2
2 x n x
L
2
Putting n 2 , Bn
an 0
sin
L
sin
L
dx
4 x
1 cos
2 2 x L dx 2 L L
L L
2 2
2 0 2 0
sin dx
L 2 2 2 2
Q34. The solution of the differential equation
dx
x2
dt
with the initial condition x0 1 will blow up as t tends to
1
(a) 1 (b) 2 (c) (d)
2
Ans: (a)
dx dx x 21 1
Solution: x 2 2 dt t C t C
dt x 2 1 x
1 1 1
x0 1 0 C C 1 t 1 x as t 1 , x blows up
1 x 1 t
1
Q35. The inverse Laplace transforms of is
s s 1
2
1 2 t 1 2
(a) t e (b) t 1 e t
2 2
t 1 e t
1 2
(c) t 1 e t (d)
2
Ans: (c)
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1 t t
Solution: f s
1
f t e t L1 e dt e t 0 e t 1
t
s 1 ss 1 0
1 t
e 1 dt e t
t
L1 2 t t
et t 1 .
s s 1 0 0
Q36. The approximation cos 1 is valid up to 3 decimal places as long as is less than:
NET/JRF -(DEC-2013)
Q37. If A iˆyz ˆjxz kˆxy , then the integral A dl (where C is along the perimeter of a
C
(a)
1 3
2
a b3 (b) ab 2 a 2 b (c) a 3 b 3 (d) 0
Ans: (d)
A d l A.d a 0 since A 0 .
C S
Q38. If A, B and C are non-zero Hermitian operators, which of the following relations must
be false?
(a) A, B C (b) AB BA C (c) ABA C (d) A B C
Ans: (a)
Solution: A, B C AB BA C ( AB BA)† C †
(( AB)† ( BA)† ) C † ( B † A† ) ( A† B † ) C †
Hence A,B and C are hermitian then
BA AB C A, B C
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Q39. Which of the following functions cannot be the real part of a complex analytic function
of z x iy ?
z 3 x iy x 3 iy 3 3ixy x iy
3
Ans: (b)
1 2 x1
Solution: 2
2
x1 x1 x 2 x3 x 4 x1 x2 x3 x4
2 2 2 2 2 2 2
2
2
2 2 2
2 x
1 x2 x3 x4 1 2 2 x1 x1 x12 x22 x32 x42
2
x12
4
x12 x22 x32 x42
2 3 4 1 1 2 3
x 2 x 2 x 2 x 2 2 4 x 2 8x 2 2 x 2 x 2 x 2 x 2
2 1 1 4
x1 x 2 x3 x 4
2 2 2 2 3
x1 x 2 x3 x 4
2 2 2 2 3
Now similarly solving all and add up then we get
2 2 2 2 1 1 1 1
2 2 2 2 2 2 2 2
x1 x 2 x3 x 4 x1 x 2 x3 x 4
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2 2
8 x1 x2 x3 x4 8 x1 x22 x32 x42
2 2 2
0
3
x12 x22 x32 x42
(a) (b) 3
(c) (d)
y 3
4y 8y 3
2 y3
Ans: (b)
1
Solution: y
dx
dx
, pole is of 2nd order at x iy , residue 1/ 4iy 3
x2
2 2
0
2 2 y x
2 2
1 1
Integral 2 i 3
2 4iy 4 y3
Q42. The Fourier transform of the derivative of the Dirac - function, namely x , is
proportional to
(a) 0 (b) 1 (c) sin k (d) ik
Ans: (d)
Solution: Fourier transform of x
H K x e ikx dx ike k 0 ik
Q43. Consider an n nn 1 matrix A , in which Aij is the product of the indices i and j
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1 2
Solution: If matrix is 2 2 let then eigen value is given by
2 4
1 2
0 (1 )(4 ) 4 0 0,5
2 4
1 2 3
If If matrix is 3 3 let 2 4 6 then eigen value is given by
3 6 9
1 2 3
2 4 6 0
3 6 9
100
C
B
10
1
A
0.1 1 10 100 1000
0.1
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NET/JRF -(JUNE-2014)
Q45. Consider the differential equation
d 2x dx
2
2 x0
dt dt
with the initial conditions x0 0 and x 0 1 . The solution xt attains its maximum
value when t is
(a) 1/2 (b) 1 (c) 2 (d)
Ans: (b)
d 2x dx
2 x 0 m 2 2m 1 0 m 1 0 m 1, 1
2
Solution: 2
dt dt
x c1 c2 t e t , since x 0 0 0 c1 x c2te t
x c2 te t e t
Since x 0 1 1 c2 x te t
e t 0, 1 t 0 t , t 1
x e t 1 1 t e t 1 e t t 1 e t
x 1 e 1 0e t 0
Q46. Consider the matrix
0 2i 3i
M 2i 0 6i
3i 6i 0
The eigenvalues of M are
(a) 5, 2, 7 (b) 7, 0, 7 (c) 4i, 2i, 2i (d) 2, 3, 6
Ans: (b)
0 2i 3i 0 2i 3i
Solution: M 2i 0 6i , M 2i 0 6i
3i 6i 0 3i 6i 0
M M
Matrix is Hermitian so roots are real and trace = 0.
1 2 3 0, 1 2 3 0 7, 0, 7
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1
Q47. If C is the contour defined by z , the value of the integral
2
dz
C sin 2 z
is
(a) (b) 2 i (c) 0 (d) i
Ans: (c)
1 1
Solution: f z 2 z
sin z 2
3 5
z z 1 1
sin z z .... 2
2
3 5 sin z z3 z5
z ....
3 5
2
1 1 z2 z4 dz
2
1 .... 0
2
sin z z 3 5 C sin 2 z
Q48. Given
n 0
Pn x t n 1 2 xt t 2 1 / 2
, for t 1 , the value of P5 1 is
k k
(c) Re f k (d) Re f k
k k
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Ans: (b)
Solution: This is cosine function
A
f x A cos x F k k k0 k k0
2
NET/JRF (DEC-2014)
Q50. Let r denote the position vector of any point in three-dimensional space, and r r .
Then
(a) r 0 and r r / r (b) r 0 and 2 r 0
(c) r 3 and 2 r r / r 2 (d) r 3 and r 0
Ans: (d)
Solution: r xxˆ yyˆ zzˆ
x y z
r 111 3
x y z
xˆ yˆ zˆ
z y x z y x
r / x / y / z xˆ yˆ zˆ 0
x y z y z z x x y
a 0 0 1
Q51. The column vector b is a simultaneous eigenvector of A 0 1 0 and
a 1 0 0
0 1 1
B 1 0 1 if
1 1 0
(a) b 0 or a 0 (b) b a or b 2a
(c) b 2a or b a (d) b a / 2 or b a / 2
Ans: (b)
Solution: Let b a
0 0 1 a a 0 1 1 a a
0 1 0 a a and 1 0 1 a 2 a
1 0 0 a a 1 1 0 a a
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Let b 2a
0 0 1 a a 0 1 1 a a a
0 1 0 2a 2a and 1 0 1 2a 2a 1 2a
1 0 0 a a 1 1 0 a a a
For other combination above relation is not possible.
sin 2 x
Q52. The principal value of the integral x3
dx is
z 0 z 0 z
1 d2 3 e
i2z
Residue R lim 2 z 0 3
2
2! z 0 dz z
f x dx iR i 2 2 i Im. Part 2 f x dx 2
zn n 1 1
n n! for all z
(a) (b) n 0
z n only if 0 z 1
z n!
1 1 zn
n0 z n z n n n! only if z 1
(c) for all 0 z (d)
n!
Ans: (c)
z2 zn 1 1 1
1
Solution: e 1 z ....
z
and e 1
1/ z
2
.... n
2! n 0 n ! z 2! z n 0 z n !
11
f z e z e1/ z z n
, for all 0 z
n 0 z n n!
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Q54. Two independent random variables m and n , which can take the integer values
0, 1, 2, ..., , follow the Poisson distribution, with distinct mean values and
respectively. Then
(a) the probability distribution of the random variable l m n is a binomial distribution.
(b) the probability distribution of the random variable r m n is also a Poisson
distribution.
(c) the variance of the random variable l m n is equal to
(d) the mean value of the random variable r m n is equal to 0.
Ans: (c)
Solution: l2 m2 n2
1
Q55. Consider the function f z
ln1 z of a complex variable
z
z re i r 0, . The singularities of f z are as follows:
(a) branch points at z 1 and z ; and a pole at z 0 only for 0 2
(b) branch points at z 1 and z ; and a pole at z 0 for all other than 0 2
(c) branch points at z 1 and z ; and a pole at z 0 for all
(d) branch points at z 0, z 1 and z .
Ans: None of the above is correct
1 1 z 2 z3 z z2
Solution: For f z ln 1 z z ..... 1 .....
z z 2 3 2 3
There is no principal part and when z 0 , f z 1 . So there is removable singularity
1 2 n 1
n
x
Q56. The function f x , satisfies the differential equation
n 0 n ! n 1 ! 2
(a) x
d2 f
2
dx 2
x
df
dx
x2 1 f 0 (b) x
d2 f
dx
2
2
2x
df
dx
x2 1 f 0
(c) x
d2 f
2
dx 2
x
df
dx
x2 1 f 0 (d) x
d2 f
dx
2
2
x
df
dx
x2 1 f 0
Ans: (c)
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1 x
2 n 1
n
Solution: f x is generating function (Bessel Function of first kind)
n 0 n ! n 1 ! 2
d2 f df
x x 2 n 2 f 0 , put n 1 .
which satisfies the differential equation x 2
dx 2
dx
Q57. Let and be complex numbers. Which of the following sets of matrices forms a
group under matrix multiplication?
1
(a) (b) , where 1
0 0
* 2 2
(c) *
, where * is real (d) , where
*
1
*
Ans: (d)
2 2
Solution: 1
* *
x , p , Lk (where i j k is the Levi-Civita symbol, x , p, L are
3
Q58. The expression i jk i j
i , j , k 1
the position, momentum and angular momentum respectively, and A, B represents the
Poisson Bracket of A and B ) simplifies to
(a) 0 (b) 6
(c) x , p L
(d) x p
Ans: (b)
NET/JRF (JUNE-2015)
dx
Q59. The value of integral 1 x 4
(a) (b) (c) 2 (d) 2
2 2
Ans. (a)
dz
Solution: 1 z4
z R
2 n 1
Now, pole z e 4
i
1 1 1 1
n 0, z0 e 4 i , n 2 z2 i
2 2 2 2
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i 3
1 1 1 1
n 1 z1 e 4 i , n 3 z3 i
2 2 2 2
only z0 and z1 lies in contour
i
1 1 1
i.e., residue at z e 4 i
4 2 2
i 3
1 1 1
residue at z e 4 i
4 2 2
dx
now x 1
4
2 i Re S
2
d 2x dx
Q60. Consider the differential equation 2
3 2 x 0 . If x 0 at t 0 and x 1 at t 1 ,
dt dt
the value of x at t 2 is
(a) e 2 1 (b) e 2 e (c) e 2 (d) 2e
Ans. (b)
Solution: D 2 3D 2 0
D 1 D 2 0 D 1, 2 x c1e2t c2et
using boundary condition x 0, t 0 c1 c2
1 1 e2t 1 t
c2 , c1 x e
ee 2
e e
2
e e e e2
2
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a
L sin at 2
s a2
6 4 3 4 36 12
L 6t 3 3sin 4t 2 4 2
s 4
s 16 s s 16
2 f 2 f
2
Q62. Let f x, t be a solution of the wave equation v in 1 -dimension. If at
t 2 x 2
f
t 0, f x, 0 e x and x, 0 0 for all x , then f x, t for all future times t 0 is
2
t
described by
x 2 v 2t 2
(b) e x vt
2
(a) e
1 x vt 2 3 x vt 2 1 x vt 2 x vt
2
(c) e e (d) e e
4 4 2
Ans. (d)
2 f 2 f
2
Solution: For v
t 2 x 2
f
x, 0 0 and f x, 0 e x
2
t
1 1
2
f x vt f x vt . Therefore, solution is f e x vt e x vt
2
f
2 2
NET/JRF (DEC-2015)
Q63. In the scattering of some elementary particles, the scattering cross-section is found to
depend on the total energy E and the fundamental constants h (Planck’s constant) and c
(the speed of light in vacuum). Using dimensional analysis, the dependence of on
these quantities is given by
2
hc hc hc hc
(a) (b) 3 / 2 (c) (d)
E E E E
Ans.: (c)
Solution: The dimension of is dimension of “Area”
h Joul sec
c m / sec
E Joul
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2
hc
m dimension of area
2
E
1
Q64. If y , then x is
tanh x
y 1 y 1 y 1 y 1
(a) ln (b) ln (c) ln (d) ln
y 1 y 1 y 1 y 1
Ans.: (d)
1
Solution: y
tanh x
e x e x e2 x 1
y
e x e x e2 x 1
ye 2 x y e 2 x 1 ye 2 x e 2 x 1 y e 2 x y 1 1 y
1
y 1 1 y 1 y 1 2
2 x ln x ln ln
y 1 2 y 1 y 1
z
Q65. The function of a complex variable z has
sin z 2
(a) a simple pole at 0 and poles of order 2 at n for n 1, 2,3...
2 z
lim z n . , exists. So its pole of order 2 .
z n sin z 2
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Q66. The Fourier transform of f x is f k dxeikx f x .
(a) i k i k 2 (b) k k 2
(c) i k k 2 (d) i k i k 2
Ans.: (c)
Solution: f k dx e x x x
ikx
x e dx
ikx
x eikx dx eikx x ike x dx i k
ikx
x e dx k 2
ikx
dx
Q67. The solution of the differential equation 2 1 x 2 , with initial condition x 0 at
dt
t 0 is
sin 2t , 0 t 4 sin 2t , 0 t 2
(a) x (b) x
sinh 2t , 1,
t t
4 2
sin 2t , 0 t 4
(c) x (d) x 1 cos 2t , t 0
1,
t
4
Ans.: (c)
dx dx
Solution: 2 1 x2 , 2dt , sin 1 x 2t c , x 0, t 0 so, c 0 x sin 2t
dt 1 x 2
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1 sin 2t , sin sin 2t , t .
2 4
sin 2t , 0t
So, x 4
1
, t
4
Q68. The Hermite polynomial H n x , satisfies the differential equation
d 2Hn dH n
2
2x 2nH n x 0
dx dx
The corresponding generating function
1
G t, x H n x t n , satisfies the equation
n 0 n !
2G G G 2G G G
(a) 2x 2t 0 (b) 2x 2t 2 0
x 2
x t x 2
x t
2G G G 2G G 2G
(c) 2x 2 0 (d) 2 x 2 0
x 2
x t x 2 x xt
Ans.: (a)
1 1 1
Solution: G H n x t n , G H n x t n , G H n x t n
n! n! n!
G 1
H n x n t n 1
t n!
Let’s check the options one by one
G G G
2x 2t 0
x 2
x t
1 1 1
H n x t n 2 x H n x t n 2t H n x nt n 1
n! n! n!
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Ans.: (a)
2 1
y0 4 y1 y2 2 y3 y4
x
Solution: I y
3 x 5
2
2 1 1 1 1 1 1
4 2 0
3 5 9 4 21 69 5
2 1
2 1
0.5734 0.09523 0.0145 9
3 5 4 1
2 21
0.2 0.5734 0.09523 0.0145 1
3 6
31
2 8 1
0.8831 0.5887
3 69
Q70. Consider a random walker on a square lattice. At each step the walker moves to a nearest
neighbour site with equal probability for each of the four sites. The walker starts at the
origin and takes 3 steps. The probability that during this walk no site is visited more than
one is
(a) 12 / 27 (b) 27 / 64 (c) 3 / 8 (d) 9 /16
Ans.: (d)
Solution: Total number of ways 4 4 4
Number of preferred outcome 4 3 3
( Any four option in step-1 and only 3 option in step 2 &3 because he can not go to
previous position)
4 3 3 9
probability
4 4 4 16
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NET/JRF (JUNE-2016)
1
Q71. The radius of convergence of the Taylor series expansion of the function
cosh x
around x 0 , is
(a) (b) (c) (d) 1
2
Ans: (c)
1 e4 z 1
Q72. The value of the contour integral cosh z 2sinh z dz around the unit circle C
2 i
C
f z
2e z e 4 z 1 2 e e
5z z
3 e 2z
3 e 2z
ln 3
For pole at z z0 ,3 e 2 z0 0 e 2 z0 3 z0
2
It has simple pole at z0
Re z0 lim z z0 f z lim z z0
2 e5 z e z
z z0 z z0 3e 2z
z z0 2 5e5 z e z 2 e5 z e z 1 e5 z0 e z0
lim
z z0 2e 2 z e
2 z0
3 3
5
9 3 3 8
3 3 3
1 1 8
f z dz 2 i Residue
2 i 2 i 3
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Q73. The Gauss hypergeometric function F a, b, c, z , defined by the Taylor series expansion
around z 0 as F a, b, c, z
a a 1 ... a n 1 b b 1 ... b n 1
n 0 c c 1 ... c n 1 n !
zn ,
dF ab
F a 1, b 1, c 1, z
dz c
Q74. Let X and Y be two independent random variables, each of which follow a normal
distribution with the same standard deviation , but with means and ,
respectively. Then the sum X Y follows a
(a) distribution with two peaks at and mean 0 and standard deviation 2
(b) normal distribution with mean 0 and standard deviation 2
(c) distribution with two peaks at and mean 0 and standard deviation 2
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Ans: (d)
Solution: x y 0
12 x2 y2 2 2
2
Q75. Using dimensional analysis, Planck defined a characteristic temperature TP from powers
hc 5 hc3 G hk B2
(a) (b) 2 (c) (d)
k B2G kBG hc 4 k B2 Gc 3
Ans: (a)
E ML2T 2
Solution: E h h ML2T 1
T 1
E ML2T 2
E k BT k B ML2T 2TP1
T TP
m1m2 MLT 2 L2
F G 2 G 2
G M 1 L3T 2
r M
5
hc 5 ML2T 1 LT 1 ML7T 6
TP2 TP
k B2G M 2 L4T 4TP2 M 1 L3T 2 7 6 2
ML T TP
1 1 1 1
(a) (b) (c) (d)
1 i k 1 i k k i k i
Ans: (b)
dn dn
Solution: f x x n x n x n x
n 1 dx n 0 dx n 0
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f x n x
n 0
1 1
F f x ik 1 ik ik ik ....
n 2 3
n 0 1 ik 1 ik
Q77. In finding the roots of the polynomial f x 3 x 3 4 x 5 using the iterative Newton-
Raphson method, the initial guess is taken to be x 2 . In the next iteration its value is
nearest to
(a) 1.671 (b) 1.656 (c) 1.559 (d) 1.551
Ans: (b)
Solution: f x 3 x 3 4 x 5 ; f x 9 x 2 4
3xn3 4 xn 5 3 x03 4 x0 5
xn 1 xn x1 x0
9 xn2 4 9 x02 4
3 8 4 2 5 11
Let x0 2 x1 2 2 x1 1.656
9 4 4 32
NET/JRF (DEC-2016)
1 3 2
Q78. The matrix M 3 1 0 satisfies the equation
0 0 1
(a) M 3 M 2 10M 12 I 0 (b) M 3 M 2 12 M 10 I 0
(c) M 3 M 2 10M 10 I 0 (d) M 3 M 2 10 M 10 I 0
Ans. : (c)
Solution: The characteristic equation is
1 3 2
3 1 0 0
0 0 1
1 1 1 3 3 1 0
2 1 1 9 1 0 3 10 2 10 0
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M M 10 M 10 I 0 then the correct option is (c)
3 2
is
1 e sT 1 e sT
1 e sT
1 e sT
t t T T
L f t L L uT t L uT t
T T
1 e sT 1 T e sT 1 e sT
2
s 2T T s2 s s sT
1
Q80. The Fourier transform dxf x eikx of the function f x is
x 2
2
2k 2k
(a) 2 e (b) 2 e 2k
(c) e 2k
(d) e
2 2
Ans. : (d)
1 1 a k
Solution: Fourier transform of f x , a 0 is x eikx dx e
x a2
2 2
a 2
a
1
x
2k
Hence eikx dx e
2
a 2
2
Q81. Given the values sin 450 0.7071, sin 500 0.7660, sin 550 0.8192 and sin 600 0.8660 ,
the approximate value of sin 520 , computed by Newton’s forward difference method, is
(a) 0.804 (b) 0.776 (c) 0.788 (d) 0.798
Ans. : (c)
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Solution: Given -
x y f x f x 2 f x 3 f x
450 0 7071
h 5 , 520 x0 uh 450 uh
52 45 52 45 7
u
h 5 5
From table we have
y0 0 7071, y0 0 0589 , 2 y0 0 0057
3 y0 0 0007 , 4 y0 0,.....
y0 2 y0 3 y0
sin 520 y0 u u u 1 u u 1 u 2
1 2 3
7 0 0057 7 7 0 0007 7 7 7
0 7071 0 0589 1 1 2 0
5 2 55 6 5 5 5
0 0399 0 0049
0 7071 0 0825
25 125
0 7071 0 0825 0 0016 0 0000 0 7880
f 2 f
Q82. Let f x, t be a solution of the heat equation D 2 in one dimension. The initial
t x
condition at t 0 is f x, 0 e x for x . Then for all t 0, f x, t is given by
2
2
[Useful integral: dx e x ]
2 2
1
x
1
x
(a) e 1 Dt (b) e 1 2 Dt
1 Dt 1 2 Dt
2
x2
1
x
(c) e 1 4 Dt (d) e 1 Dt
1 4 Dt
Ans. : (c)
f 2 f
Solution: D 2 , t 0
t x
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f f
2
Let D c 2 , c2 2
t x
Initial condition f x , 0 e x g x , x
2
2
x 2 cz t
Now, g x 2cz t e
x 2
2
e x 4 c 2 z 2t 4 cxz t z 2 dz e
1 4 c 2t z 2 4 cxz t
e
e
dz
exp 1 4c t z 2 z 1 4c 2t 1 4c 2t 1 4c 2t dz
cx t
2 2
2 cx t
2
4 c 2 x 2t 2 cx t
2
e x
2
1 4 c 2t z 2 1 4c t dz e 2 x2 4 c 2 x 2t
1 4 c 2t z 2
e
1 4 tc
e
e 1 4 c 2t
e
1 4 tc
dz
x2
e 1 4 c 2t
1
2
2 1 4c 2t
2
x
1
f x, t
1 4 Dt
e1 4 Dt c 2
D
3
Q83. A stable asymptotic solution of the equation xn 1 1 is x 2 . If we take
1 xn
n 1
xn 2 n and xn 1 2 n 1 , where n and n1 are both small, the ratio is
n
approximately
1 1 1 2
(a) (b) (c) (d)
2 4 3 3
Ans. (c)
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NET/JRF (JUNE-2017)
Q84. Which of the following can not be the eigen values of a real 3 3 matrix
(a) 2i, 0, 2i (b) 1,1,1 (c) ei , e i ,1 (d) i, 1, 0
Ans. : (d)
Solution: If the matrix is real then the complex eigen values always occurs with its complex
conjugate. In option (d) if i is an eigen value then i must also be an eigen value. But
i is not given in option, hence option (d) is incorrect.
Q85. Let u x, y e ax cos by be the real part of a function f z u x, y iv x, y of the
v a e sin by
2 ax
(iv)
x b
From equation (iii) and (iv)
a 2 e ax sin by
be sin by
ax
b
b2 a 2 b a
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i z / 2
ze
Q86. The integral 2 dz along the closed contour shown in the figure is
z 1
y
1 1 x
For z 1
I 2 i lim
zei z / 2
2 i
1 e i / 2
ie i / 2
z 1 z 1 2
Integral i
e i / 2
e i / 2
2i 2 i 2 sin
2
2i 2
dy cos x
Q87. The function y x satisfies the differential equation x 2y . If y 1 1 , the
dx x
value of y 2 is
cos x sin x c
Hence y.x 2 x 2 . dx c y 2
x 2
x2 x
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sin x 1
when x 1, y 1 hence c 1 y 2
x2 x
1
hence, when x 2, y
4
Q88. The random variable x x is distributed according to the normal distribution
x2
1
P x e 2 2
. The probability density of the random variable y x 2 is
2 2
1 2 1 2
(a) e y / 2 , 0 y (b) e y / 2 , 0 y
2 y 2
2 2 y 2
1 2 1 2
(c) e y / 2 , 0 y (d) e y / , 0 y
2 2
2 y 2
Ans. : (a)
x2
1
Solution: p x e 2 2
, x
2 2
x2
1
p x dx 1
2 2
e 2 2
dx 1
x2 x2
1 2 1
2 e 2 2
dx 1 e 2 2
dx 1
0 2 2 2 2 0
1
put x 2 y dy 2 xdx , dx
2 y
y y
1 1 1 1 2 2
2
2 2
e
0
2 2
2 y
dy
2 2
0 y
e dy
y
1
f y e 2 2
, 0 y
2 y 2
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Q89. The Green’s function satisfying
d2
g x, x0 x x0
dx 2
with the boundary conditions g L, x0 0 g L, x0 , is
1 1
2 L x0 L x L , L x x0 2 L x0 L x L , L x x0
(a) (b)
1 x L x L , x0 x L 1 x L x L , x0 x L
2 L 0 2 L 0
1
2 L L x0 x L , L x x0
1
(c) (d) x L x L , L x L
1 x L L x , x0 x L
2L
2 L 0
Ans. : (a)
d2
Solution: 2 g x, x0 x x0
dx
boundary conditions:
g L, x0 0 g L, x0
g L, x0 0 CL D 0 CL D
A x L , x x0
Hence, g x, x0
C x L , x x0
A x0 L C x0 L
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x L
AC 0
x0 L
g
From discontinuity of derivative of Green’s function at x x0
x
We have
g g
1
x x x0 x x x0
g g
C, A
x x x0 x x x0
C A 1 C A 1
Thus, the required solution of Green’s function is given by
x0 L x L
, x x0
g x, x0 2 L
x0 L x L , x x0
2L
i z
Q90. Let x , y , z be the Pauli matrices and x x y y z z exp
2
i z
x x y y z z exp 2 .
Then the coordinates are related as follows
x cos sin 0 x x cos sin 0 x
(a) y sin cos 0 y (b) y sin cos 0 y
z 0 0 1 z z 0 0 1 z
cos 2 sin 0 cos 2 sin 0
x 2 x 2
x x
(c) y sin cos 0 y (d) y sin cos 0 y
2 2 2 2
z z z z
0 0 1 0 0 1
Ans. : (b)
0 1 0 i 1 0
Solution: x , y and z
1 0 i 0 0 1
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z x iy
Hence, x x y y z z
x iy z
z' x1 iy '
x ' x y ' y z ' z
x ' iy ' z '
i / 2 i / 2
i z e 0 i z e 0
exp
i / 2
and exp
z 0 e 2 0 ei / 2
z x iy z ei x iy
x iy z e i x iy z
Hence, z z and x iy ei x iy
x cos sin 0 x
Thus, y sin cos 0 y
z 0 0 1 z
Q91. Which of the following sets of 3 3 matrices (in which a and b are real numbers) forms
a group under matrix multiplication?
1 0 a 1 a 0
(a) 0 1 0 ; a, b (b) 0 1 b ; a, b
b 0 1
0 0 1
1 0 a 1 a 0
(c) 0 1 b ; a, b (d) b 1 0 ; a, b
0 0 1
0 0 1
Ans. : (c)
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Solution: In order to form the group the required matrix must satisfy the following conditions.
(a) For any three matrices A, B, C G A BC AB C
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CLASSICAL MECHANICS SOLUTIONS
NET/JRF (JUNE-2011)
b
Q1. A particle of unit mass moves in a potential V x ax 2 , where a and b are positive
x2
constants. The angular frequency of small oscillations about the minimum of the potential
is
(a) 8b (b) 8a (c) 8a / b (d) 8b / a
Ans: (b)
1
b V 2b b 4
Solution: V x ax 2 2 0 2ax 3 0 ax 4 b 0 x0 .
x x x a
k 2V
Since , m 1 and k 2 where x0 is stable equilibrium point.
m x x x0
1
2V 6b 6b b 4
Hence k 2 2a 4 2a 8a at x x0 .
x x0 b a
a
Thus, 8a .
Q2. The acceleration due to gravity g on the surface of Earth is approximately 2.6 times
that on the surface of Mars. Given that the radius of Mars is about one half the radius of
Earth, the ratio of the escape velocity on Earth to that on Mars is approximately
(a) 1.1 (b) 1.3 (c) 2.3 (d) 5.2
Ans: (c)
Solution: Escape velocity = 2 gR
Escape velocity of Earth ge R e R g
2.3 where e 2 and e 2.6.
Escape velocity of Mass gm R m Rm gm
Q3. The Hamiltonian of a system with n degrees of freedom is given by H q1 ,.....q n ; p1 ,....... p n ; t ,
Q5. Consider the decay process in the rest frame of the . The masses of the
A. The energy of is
(a)
M 2
M 2 c 2
(b)
M 2
M 2 c 2
(c) M M c 2 (d) M M c2
2M 2M
Ans. : (b)
Solution:
From conservation of energy M c2 E E .
M 2 c 4
M c 2 E E , M 2 c 4 E2 E2 E E
M c 2
E E
M 2 c 2
and E E M c E
2
M2 M 2 c 2
.
M 2M
B. The velocity of is
(a)
M M 2 c
2
(b)
M M 2 c
2
(c)
M c
(d)
M c
M 2 M 2 M 2 M 2 M M
Ans: (a)
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Solution: Velocity of
E
M 2
M 2 c 2
M c 2 v2
1 2
4 M 2 M 2
2M c M 2 M 2
2
v2
1
c2
v2 4M 2 M 2 v 2 M 4 M 4 2 M 2 M 2 4 M 2 M 2 M 2 M 2
1 v 2 2
c.
c2 M 2 M 2 c M 2 M 2 M M
2 2 2
Q6. The Hamiltonian of a particle of unit mass moving in the xy -plane is given to be:
1 2 1 2
H xp x yp y x y in suitable units. The initial values are given to be
2 2
x0, y0 1,1 and p x 0, p y 0 1 , 1 . During the motion, the curves traced out
2 2
by the particles in the xy-plane and the p x p y – plane are
After solving these four differential equation and eliminating time t and using boundary
1 1 1
condition one will get x and px
y 2 py
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NET/JRF (DEC-2011)
Q7. A double pendulum consists of two point masses m attached by strings of length l as
shown in the figure: The kinetic energy of the pendulum is
(a)
2
1 2 2 2
ml 1 2
l
(b)
2
ml 21 2 212 cos1 2
1 2 2 2
1
(c)
1 2 2
2
ml 1 222 212 cos1 2 m
l
(d)
ml 2 1 2 212 cos 1 2
1 2 2 2
2
2
Ans: (b) m
x 2 l cos 11 l cos 2 2 , y 2 l sin 11 l sin 2 2
Put the value of x1 , y1 , x 2 , y 2 in K.E equation, one will get
1 2 2 2
T ml 21 2 212 cos 1 2 .
2
Q8. A constant force F is applied to a relativistic particle of rest mass m. If the particle starts
from rest at t = 0, its speed after a time t is
(a) Ft / m
Ft
(b) c tanh
(c) c 1 e Ft / mc (d)
Fct
mc F 2t 2 m 2c 2
Ans: (d)
dp
Solution: F p Ft c . At t 0, p 0 so, c 0
dt
F
mu t Fct
p Ft Ft u m .
u2 Ft
2
F 2 t 2 m2 c 2
1 2 1
c mc
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Q9. The potential of a diatomic molecule as a function of the distance r between the atoms is
a b
given by V r 12 . The value of the potential at equilibrium separation between
r 6
r
the atoms is:
(a) 4 a 2 / b (b) 2 a 2 / b (c) a 2 / 2b (d) a 2 / 4b
Ans: (d)
a b V 1 12b
Solution: V r
a 12b
, for equilibrium 0 6 0 6 a 0
r 6 r 12 r r 7
r 13
r
7
r 6
1 1
12b 12b 6 2b 6
6a 6 0 r r
r 6a a
1
2b a b a2 a2 a2
V r
6
.
a 2b 2b 2 2b 4b 4b
a a
Q10. Two particles of identical mass move in circular orbits under a central potential
V r
1 2
kr . Let l1 and l2 be the angular momenta and r1 , r2 be the radii of the orbits
2
l1 r
respectively. If 2 , the value of 1 is:
l2 r2
Q11. A particle of mass m moves inside a bowl. If the surface of the bowl is given by the
equation z
1
2
a x 2 y 2 , where a is a constant, the Lagrangian of the particle is
(a)
1
2
m r 2 r 2 2 gar 2 (b)
1
2
m 1 a 2 r 2 r 2 r 2 2
(c)
1
2
m r 2 r 2 2 r 2 sin 2 2 gar 2 1
(d) m 1 a 2 r 2 r 2 r 2 2 gar 2
2
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Ans: (d)
Solution: L
1
2
1
m x 2 y 2 z 2 mgz , where z a x 2 y 2 .
2
a r2 .
1
It has cylindrical symmetry. Thus x r cos , y r sin , z
2
x r cos r sin , y r sin r cos and z a rr .
1 2 b2 a 2 b a 1 b 1
mv1 GMm mv12 GMm
b ab a b a
2
2 2
1 2 GMm b 1 GMm
E mv1 GMm
2 a a b a a
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Q13. An annulus of mass M made of a material of uniform density has inner and outer radii a
and b respectively. Its principle moment of inertia along the axis of symmetry
perpendicular to the plane of the annulus is:
1 b 4 a 4
(a) M 2
2 b a 2
(b)
1
2
M b 2 a 2
(c) M b 2 a 2
1 1
(d) M b 2 a 2
2 2
Ans: (d)
Q14. The trajectory on the zpz - plane (phase-space trajectory) of a ball bouncing perfectly
elastically off a hard surface at z = 0 is given by approximately by (neglect friction):
PZ PZ
(a) (b)
z z
PZ PZ
(c) (d)
z z
Ans: (a)
Pz2 P2
Solution: H mgz and E z mgz .
2m 2m
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NET/JRF (JUNE-2012)
Q15. The bob of a simple pendulum, which undergoes small oscillations, is immersed in water.
Which of the following figures best represents the phase space diagram for the pendulum?
p p
(a) (b)
x x
p p
(c) (d)
x x
Ans: (d)
Solution: When simple pendulum oscillates in water it is damped oscillation so amplitude
continuously decrease and finally it stops.
Q16. Two events separated by a (spatial) distance 9 109 m , are simultaneous in one inertial
frame. The time interval between these two events in a frame moving with a constant
speed 0.8 c (where the speed of light c 3 108 m / s ) is
(a) 60 s (b) 40 s (c) 20 s (d) 0 s
Ans: (b)
Solution: x2' x1' 9 10 9 m and t 2' t1' 0 . Then
t ' v x' t1 v x'
2 c2 2
t 2 t1
1 1 ' '
' '
c 2 t t t 2 t1 v x 2 x1 v x 2 x1 .
' '
2
2 1
v2 c c2
2
1 v
2
1 v v2 v2
1 1 1
c2 c2 c2 c2 c2
Put v 0.8c t 2 t1 40 sec
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x 2
Q17. If the Lagrangian of a particle moving in one dimensions is given by L V x the
2x
Hamiltonian is
x 2 p2
xp V x V x x V x V x
1 2 1 2
(a) (b) (c) (d)
2 2x 2 2x
Ans: (a)
L x
Solution: Since H p x x L and p x p x x p x x .
x x
px x
2
x 2 p x2 x
H p x x V x H px px x Vx H V x .
2x 2x 2
Q18. A horizontal circular platform mutes with a constant angular velocity directed
vertically upwards. A person seated at the centre shoots a bullet of mass m horizontally
with speed v. The acceleration of the bullet, in the reference frame of the shooter, is
(a) 2v to his right (b) 2v to his left
(c) v to his right (d) v to his left
Ans: (a)
Solution: Velocity of bullet = vˆj , Angular velocity= k̂ . There will be coriolis
force F 2m v .
F 2mviˆ a 2viˆ .
Q19. The Poisson bracket r , p has the value
(a) r p (b) rˆ pˆ (c) 3 (d) 1
Ans: (b)
Solution: r xiˆ yjˆ zkˆ , r x 2 y 2 z 2 , p p xiˆ p y ˆj p z kˆ ,
1/ 2
p px2 p y2 pz2
1/ 2
r p r p r p r p r p r p
r , p = x p p x y p p y z p p y
x x y y z z
x px y p y z pz rp
rˆ pˆ
r p r p r p r p
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Q20. Consider the motion of a classical particle in a one dimensional double-well
potential V x
1 2
4
2
x 2 . If the particle is displaced infinitesimally from the minimum
frequency 2
(b) the particle will execute simple harmonic motion in the right well with an angular
frequency 2
(c) the particle will switch between the right and left wells
(d) the particle will approach the bottom of the right well and settle there
Ans: (b)
Solution: V x x 2 2 V 2 x 2 2 2 x 0 x 0 , x 2 .
1 2
4 x 4
2V 2V
3 x 2
2 . At x 0 , 0 so V is maximum. Thus it is unstable point
x 2 x 2
2V
2V x 2 x x0
4 and it is stable equilibrium point with 2 1.
x 2 x 2
Q21. What is proper time interval between the occurrence of two events if in one inertial frame
events are separated by 7.5 108 m and occur 6.5 s apart?
(a) 6.50 s (b) 6.00 s (c) 5.75 s (d) 5.00 s
Ans: (b)
Solution: Proper time interval
2
6.5 6 sec.
r2 7.5
t t 2
2 2
c 3
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NET/JRF (DEC-2012)
Q22. A solid cylinder of height H, radius R and density ρ, floats vertically on the surface of a
liquid of density 0 . The cylinder will be set into oscillatory motion when a small
instantaneous downward force is applied. The frequency of oscillation is
g g g 0 g
(a) (b) (c) (d)
0 H 0 H 0H H
Ans: (d)
Solution: From Newton’s law of motion ma mg 0 Agh where A is area of cross section,
m AH .
0 gh 0 g
AHa AHg 0 Agh a 1
H H
Q23. Three particles of equal mass (m) are connected by two identical massless springs of
stiffness constant (K) as shown in the figure
K K
m m m
If x1, x2 and x3 denote the horizontal displacement of the masses from their respective
equilibrium positions the potential energy of the system is
(a)
1
2
K x12 x 22 x32 (b)
1
2
K x12 x 22 x32 x 2 x1 x3
(c)
1
2
K x12 2 x 22 x32 2 x 2 x1 x3 (d)
1
2
K x12 2 x 22 2 x 2 x1 x3
Ans: (c)
1 1
K x2 x1 K x3 x2 ,
2 2
Solution: V
2 2
2GMm 2 ab 2GMm 2 ab
(a) 2GMm 2 a b (b) 2GMm 2 a b (c) (d)
ab ab
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Ans: (d)
Solution: Assume Sun is at the centre of elliptical orbit.
1 2 GMm 1 2 GMm
Conservation of energy mv1 mv2 v2
2 a 2 b
Conservation of momentum L mv1a mv2b b
v1
s a
a
v2 v1
b
1 2 1 2 GMm GMm 1 2 2 a
2
b a
mv1 mv2 m v1 v1 2 GMm
2 2 a b 2 b ab
1 2 b2 a 2 b a 1 b 1
mv1 GMm mv12 GMm
2 b 2
ab 2 a b a
b 1
v1 2GM
a b a
b 1 2GMab 2GMm 2 ab
L mv1 a m 2GM a m L
a b a b a ab
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Hence we have to calculate L, H which is only defined into phase space i.e. p and .
p2
Then L mgl 1 cos
2ml 2
L, H L H
L H
2g
p sin and
L
0
dL
2g
p sin
p p l t dt l
Q26. Two bodies of equal mass m are connected by a massless rigid rod of length l lying in the
xy-plane with the centre of the rod at the origin. If this system is rotating about the z-axis
with a frequency ω, its angular momentum is
(a) ml 2 / 4 (b) ml 2 / 2 (c) ml 2 (d) 2ml 2
Ans: (b)
Solution: Since rod is massless i.e. M 0 .
l
Moment of inertia of the system I m1r12 m2 r22 , m1 m2 m and r1 r2
2
ml 2 ml 2 ml 2 ml 2
I I . Angular momentum, J I and J .
4 4 2 2
Q27. Which of the following set of phase-space trajectories is not possible for a particle
obeying Hamilton’s equations of motion?
(a) (b)
P P
x x
(c) (d)
P P
x x
Ans: (b)
Solution: Phase curve does not cut each other
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Q28. The muon has mass 105 MeV / c and mean life time 2.2 s in its rest frame. The mean
2
NET/JRF (JUNE-2013)
Q29. The area of a disc in its rest frame S is equal to 1 (in some units). The disc will appear
distorted to an observer O moving with a speed u with respect to S along the plane of the
disc. The area of the disc measured in the rest frame of the observer O is ( c is the speed
of light in vacuum)
1/ 2 1 / 2 1
u2 u2 u2 u2
(a) 1 2 (b) 1 2 (c) 1 2 (d) 1 2
c c c c
Ans: (a)
Solution: Area of disc from S frame is 1 i.e. a 2 1 or a a 1
u2 u2 u2
Area of disc from S frame is a b a a 1 1 1 1
c2 c2 c2
u2
where b a 1 .
c2
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Q30. A planet of mass m and an angular momentum L moves in a circular orbit in a potential,
V r k / r , where k is a constant. If it is slightly perturbed radially, the angular
frequency of radial oscillations is
L2 k L2 k
2 . Thus r r0 ,
mr 3
r mk m
r
d 2Veff 3L 2
2k 3L 2
2k 3m k 3
2m k
4
m k43 4 3
k 3 6
dr 2
mr 4
r r r0 L2
4
L2
3
L6 L6 L
r r0
m
mk mk
d 2V
dr 2 r r0 mk 2
.
m L3
Q31. The number of degrees of freedom of a rigid body in d space-dimensions is
(a) 2d (b) 6 (c) d d 1 / 2 (d) d!
Ans: (c)
Q32. A system is governed by the Hamiltonian
1
p x ay 2 1 p x bx 2
H
2 2
where a and b are constants and p x , p y are momenta conjugate to x and y respectively.
a 2, b 3
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Q33. The Lagrangian of a particle of mass m moving in one dimension is given by
1 2
L mx bx
2
where b is a positive constant. The coordinate of the particle xt at time t is given by: (in
following c1 and c 2 are constants)
b 2
(a) t c1t c 2 (b) c1t c 2
2m
bt bt bt bt
(c) c1 cos c 2 sin (d) c1 cosh c 2 sinh
m m m m
Ans: (a)
d L L d
Solution: Equation of motion 0 mx b 0 mx b 0 mx b
dt x x dt
d 2x b dx b b t2
t c1 x c1t c2
dt 2 m dt m m 2
NET/JRF (DEC-2013)
Q34. Let A, B and C be functions of phase space variables (coordinates and momenta of a
mechanical system). If , represents the Poisson bracket, the value of
A, B, C A, B, C is given by
(a) 0 (b) B, C , A (c) A, C , B (d) C , A, B
Ans: (d)
Solution: We know that Jacobi identity equation
A, B, C B, C , A C , A, B 0
Now A, B, C A, B, C B, C , A C , A, B
z2
Q35. A particle moves in a potential V x 2 y 2 . Which component(s) of the angular
2
momentum is/are constant(s) of motion?
(a) None (b) Lx , L y and L z (c) only L x and Ly (d) only L z
Ans: (d)
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z2
Solution: A particle moves in a potential V x 2 y 2
2
r 2
V r , , r 2 sin 2 cos 2 r 2 sin 2 sin 2 cos 2
2
r 2
V r , , r 2 sin 2 cos 2
2
Now is cyclic-co-ordinate p i.e Lz is constant of motion.
Q36. The Hamiltonian of a relativistic particle of rest mass m and momentum p is given
(a) L m 1 x 2 V x (b) L m 1 x 2 V x
xm
p
1 x 2
xm
Put value p L m 1 x 2 V x
1 x 2
Q37. A pendulum consists of a ring of mass M and radius R suspended by a massless rigid
rod of length l attached to its rim. When the pendulum oscillates in the plane of the ring,
the time period of oscillation is
lR 2
(a) 2
g
(b) l 2
R2 1/ 4
2 R 2 2 Rl l 2 2
(c) 2
g R l
(d) 2R 2
2 Rl l 2 1/ 4
g
Ans: (c)
Solution: The moment of inertia about pivotal point is given by
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I I c.m Md MR M (l R)
2 2 2
2 R 2 2 Rl l 2
Time period is given by 2 .
g R l
Q38. Consider a particle of mass m attached to two identical springs y
each of length l and spring constant k (see the figure). The
equilibrium configuration is the one where the springs are
o
unstretched. There are no other external forces on the system. If the
particle is given a small displacement along the x -axis, which of
the following describes the equation of motion for small x
o
oscillations?
kx 3 kx 2
(a) mx 0 (b) mx kx 0 (c) mx 2kx 0 (d) mx 0
l2 l
Ans: (a)
Solution: The lagrangian of system is given by y
1 2
L mx V ( x)
2 x
o
The potential energy is given by
2 2
k 1
k 1
2
V ( x) x 2 l 2 2
l x2 l 2 2 l
2 o x
2
1
V ( x) k x 2 l 2 2 l
For small oscillation one can approximate potential by Taylor expansion
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2
1
2
2
x
2 2
1 x 2 1 x 4
V ( x) kl 1 2 1 V ( x) kl 1 2 4 1
2
l 2 l 8 l
2
kl 2 x x 4
2
V ( x) V ( x ) k 2 .
4 l2 4l
1 2 x 4
So Lagrangian of system is given by L mx k 2
2 4l
d L L kx 3
The Lagranges equation of motion 0 m
x 0.
dt x x l2
NET/JRF (JUNE-2014)
Q39. The time period of a simple pendulum under the influence of the acceleration due to
gravity g is T . The bob is subjected to an additional acceleration of magnitude 3 g in
the horizontal direction. Assuming small oscillations, the mean position and time period
of oscillation, respectively, of the bob will be
(a) 0 o to the vertical and 3T (b) 30 o to the vertical and T / 2
g 3g 2 g 2 4g 2 2 g
3g
l l 1 T
T 2 T 2 T g
2g g 2 2 g
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1
Q40. A particle of mass m and coordinate q has the Lagrangian L mq 2 qq 2 , where
2 2
is a constant. The Hamiltonian for the system is given by
p 2 qp 2 p2
(a) (b)
2m 2 m 2 2m q
p2 qp 2 pq
(c) (d)
2m 2m q 2 2
Ans: (b)
1 2 2
Solution: H qp
L where L mq qq
2 2
L p
p mq qq p q m q q
q m q
H qp
L
p2 1
m
p2
q
p2
m q 2 m q 2 2 m q 2
p2 p2
H qp
L m q
m q 2 m q 2
p2 p2 p2
H qp
L H
m q 2 m q 2 m q
Q41. The coordinates and momenta xi , pi i 1, 2, 3 of a particle satisfy the canonical Poisson
Ans. : (d)
Solution: C1 x2 p3 x3 p2 , C2 x1 p2 x2 p1 , C3 x1 p3 x3 p1
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C C C C C C C C C C C C
C2 , C3 2 3 2 3 2 3 2 3 2 3 2 3
x1 p1 p1 x1 x2 p2 p2 x2 x3 p3 p3 x3
C2 , C3 p2 x3 x2 p3 0 x1 0 0 x1 0 p1 p2 x3 x2 p3 C1
C3 C1 C3 C1 C3 C1 C3 C1 C3 C1 C3 C1
C3 , C1
x1 p1 p1 x1 x2 p2 p2 x2 x3 p3 p3 x3
90 GeV/c2 respectively, and the decaying Higgs particle is at rest, the energy of the
photon will approximately be
(a) 35 3 GeV (b) 35 GeV (c) 30 GeV (d) 15 GeV
Ans. : (c)
Solution: H B PH Z B
From conservation of momentum 0 P1 P2 P1 P2 P1 P2
EZ B EPH E ZB
EPH M Z2B c 4 P1 P2
M Z2B c 4 M Z2B c 2
EZ B EPH EZ B EPH M H B c 2
M HB c2 M HB
2 EPH M H B c 2
M z2B c 2
EPH
M 2
HB
M z2B c 2
M HB M HB
125 125 90 90 c
4
EPH 4 30.1GeV
2 125 c
Q43. A canonical transformation relates the old coordinates q, p to the new ones Q, P by
NET/JRF (DEC-2014)
Q44. The equation of motion of a system described by the time-dependent Lagrangian
1
L e t mx 2 V x is
2
dV dV
(a) mx mx 0 (b) mx mx 0
dx dx
dV dV
(c) mx mx 0 (d) mx 0
dx dx
Ans: (a)
1 L L V t
Solution: L e t mx 2 V x e t mx and e
2 x x x
d L L d V t V t
0 e t mx t mx e t
e mxe e 0
dt x x dt x x
V t V
mx m x e 0 mx mx 0
x x
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1
V a 2
0 ax bx3 0 x a bx 2 0 x , 0
x b
2V 2V
a 3bx 2 At x 0, a (Negative so it is unstable point)
x 2
x 2
2V a
a 3b 2a (Positive so it is stable point)
x 2 b
1
a 2
x
b
2V
x 2 2a
m m
Q46. The radius of Earth is approximately 6400 km . The height h at which the acceleration
due to Earth’s gravity differs from g at the Earth’s surface by approximately 1 % is
(a) 64 km (b) 48 km (c) 32 km (d) 16 km
Ans: (c)
g 2h g 2h g 2h
Solution: 1 1 h 32 k .m.
g R g R g R
Q47. According to the special theory of relativity, the speed v of a free particle of mass m and
total energy E is:
mc 2 2 E mc 2
(a) v c 1 (b) v 1
E m E
2
mc 2 mc 2
(c) v c 1 (d) v c1
E E
Ans: (c)
2 2
v 2 mc 2
mc 2 v2 m2c 4 mc 2
Solution: E 1 2 1 v c 1
v2 c E c2 E2 E
1 2
c
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p2
Q48. The Hamiltonian of a classical particle moving in one dimension is H q 4 where
2m
is a positive constant and p and q are its momentum and position respectively. Given
that its total energy E E 0 the available volume of phase space depends on E 0 as
(a) E 03 / 4 (b) E 0
Ans: (a)
V q
p2
Solution: H q4
2m E0
Phase area p dq q
p
1 2mE0
E 4
A p dq 2mE
E0 /
1/4 E0 / 1/4
AE 1/2
E
1/4
AE 3/4 2mE0
0 0 0
p2 1
Q49. A mechanical system is described by the Hamiltonian H q, p m 2 q 2 . As a
2m 2
Q
result of the canonical transformation generated by F q, Q , the Hamiltonian in
q
the new coordinate Q and momentum P becomes
1 2 2 m 2 2 1 2 2 m 2 2
(a) Q P Q (b) Q P P
2m 2 2m 2
1 2 m 2 2 1 2 4 m 2 2
(c) P Q (d) Q P P
2m 2 2m 2
Ans: (d)
p2 1 Q
Solution: H m 2 q 2 , F F1 q, Q
2m 2 q
F1 Q
p 2 p (i)
q q
F1 1 1
P P q (ii)
Q q P
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1
From equation (i) and (ii) p QP 2 q
P
p2 1 Q2 P4 1 1 1 2 4 1
H m q
2 2
m 2 2 Q P m 2 P 2
2m 2 2m 2 P 2m 2
Q50. The probe Mangalyaan was sent recently to explore the planet Mars. The inter-planetary
part of the trajectory is approximately a half-ellipse with the Earth (at the time of launch),
Sun and Mars (at the time the probe reaches the destination)
forming the major axis. Assuming that the orbits of Earth
and Mars are approximately circular with radii RE and Sun
Earth Mars
RM , respectively, the velocity (with respect to the Sun) of RE
RM
the probe during its voyage when it is at a distance
r RE r RM from the Sun, neglecting the effect of Earth and Mars, is
R E RM R E RM r
(a) 2GM (b) 2GM
r R E R M r r R E RM
RE 2GM
(c) 2GM (d)
rRM r
Ans: (b)
Solution: Total energy E K / 2a where 2a major axis and 2a RE RM .
1 2 GMm GMm R RM r
v 2GM E
mv
2 r RE RM r RE RM
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NET/JRF (JUNE-2015)
Q51. A particle moves in two dimensions on the ellipse x 2 4 y 2 8 . At a particular instant it
is at the point x, y 2,1 and the x -component of its velocity is 6 (in suitable units).
Q52. Consider three inertial frames of reference A, B and C . the frame B moves with a
c c
velocity with respect to A , and C moves with a velocity with respect to B in the
2 10
same direction. The velocity of C as measured in A is
3c 4c c 3c
(a) (b) (c) (d)
7 7 7 7
Ans. (b)
c c A S B S C u x
Solution: v , u x
2 10
C/2 C /10
u v 4c
ux x
u vx 7
1 2
c
1 2
Q53. If the Lagrangian of a dynamical system in two dimensions is L mx mxy
, then its
2
Hamiltonian is
1 1 2 1 1 2
(a) H px p y py (b) H px p y px
m 2m m 2m
1 1 2 1 1 2
(c) H px p y py (d) H px p y px
m 2m m 2m
Ans. (c)
1 2 L
Solution: L mx mxy
mx my px (i)
2 x
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L p
mx p y or x y (ii)
y m
py px p y
put x in equation (i) p y my px y
m m
1
H px x p y y L px x p y y mx 2 mxy
2
px p y p y2
put value of x and y H
m 2m
Q54. A particle of mass m moves in the one dimensional potential V x x3 x 4 where
3 4
, 0 . One of the equilibrium points is x 0 . The angular frequency of small
oscillations about the other equilibrium point is
2
(a) (b) (c) (d)
3m m 12m 24m
Ans. (b)
V
Solution: V x x3 x4 x 2 x3 0 x0
3 4 x
2V 2 k
Spring constant k 2 ve
x x x0
m m
Q55. A particle of unit mass moves in the xy -plane in such a way that x t y t and
(a)
2
1 2
x y2 (b)
1 2
2
x y2 (c) x y (d) x y
Ans. (a)
Solution: x y and y x
x y x
y x y
and
xx 0
y y 0
and
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1
Q56. A particle moves in one dimension in the potential V k t x 2 , where k t is a time
2
d
dependent parameter. Then V , the rate of change of the expectation value V of the
dt
potential energy is
1 dk 2 k 1 dk 2 1
(a) x xp px (b) x p2
2 dt 2m 2 dt 2m
k 1 dk 2
(c) xp px (d) x
2m 2 dt
Ans. (a)
p2 1
Solution: H k t x2
2m 2
d V 1 2 p
2
1 2 x 2 k
V V , H k t x , k t x V , H
dt t 2 2m 2 2 t
d 1 xp px x 2 k x 2 k 1
V k t 2 k t xp px
dt 2 2m 2 t 2 t 2m
Q57. Let q and p be the canonical coordinate and momentum of a dynamical system. Which
of the following transformations is canonical?
1 2 1 2
1. Q1 q and P1 p
2 2
1 1
2. Q2 p q and P2 p q
2 2
(a) neither 1 nor 2 (b) both 1 and 2
(c) only 1 (d) only 2
Ans. (d)
q2 P2
Solution: For A : Q1 , P1
2 2
Q P Q P
Q1 , P1 1 . 1 1 . 1 1 (Not canonical)
q p p q
1 1
For B : Q2 p q , P2 p q
2 2
Q2 , p2 1 (canonical)
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Q58. Which of the following figures is a schematic representation of the phase space
trajectories (i.e., contours of constant energy) of a particle moving in a one-dimensional
1 2 1 4
potential V x x x p
2 4
p
(a) (b)
x
x
p p
(c) (d)
x x
Ans. (a)
x2 x4 V x
Solution: V x
2 4
V x
0 x 0, x 1
x
2V
ve for x 0 (unstable point)
x 2 E0
= + ve for x 1 (stable point)
E0 E0
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Q59. Two masses m each, are placed at the points x, y a, a and a, a and two
masses, 2m each, are placed at the points a, a and a, a . The principal moments of
Solution: I xx mi yi2 zi2 mi yi2 zi 0
i
I xz I zx 0, I yz I zy 0
(a)
k
,
2m m
3k
(b)
k
2m
13 73
5k k k 6k
(c) , (d) ,
2m m 2m m
Ans.: (a)
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1 5
Solution: L mq12 2mq22 k q12 2q22 2q1 q2
2 4
1 2 4 2 k 10 2
L mq1 mq2 q1 4q22 2q1 q2 2q2 q1
2 2 2 4
10
m 0 k 2 k
T , V 4
0 4m
2k 4k
10
4 k m 2 k 0 , 10 k 2 m 4k 4 2 m 4k 2 0
2
4
2 k 4k 4m
2
10k 2 10 2 km 4 2 km 4 4 m 2 4k 2 0
3k 2 7 2 km 2 4 m 2 0 3k 2 6 2 km 2 km 2 4 m 2 0
k 3k
k 2 2 m 3k 2 m 0
2m
,
m
Q61. Consider a particle of mass m moving with a speed v . If TR denotes the relativistic
(a) 1.25 105 (b) 5.0 105 (c) 7.5 105 (d) 1.0 104
Ans.: None of the options is correct.
1 m0 c 2
Solution: TN m0 v 2 , TR mc 2 m0 c 2 m0 c 2 ( v 0.01c )
2 v 2
1 2
c
0.01
2
1 v2
TR TN m0 v 2
T 2 2 2
Now, 1 N 1 1 1
TR TR m0 c 2 c2 1
1
m0 c 2
c 2
1 0.01
2
v 2
v 2
1 2 1 2
c c
TR TN
0.75
TR
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Q62. A canonical transformation p, q P, Q is performed on the Hamiltonian
p2 1 1
H m 2 q 2 via the generating function, F m q 2 cot Q . If Q 0 0 , which
2m 2 2
of the following graphs shows schematically the dependence of Q t on t ?
(a) (b)
Q t Q t
(c) (d)
Q t Q t
Ans.: (d)
p2 1 1
Solution: H m 2 q 2 , F1 m q 2 cot Q
2m 2 2
F1 F1 F1
p, P, KH
q Q t t
F1
p m q cot Q …..(i)
q
t
F1 1
P m q 2 cosec 2Q P
Q 2
1 P
m q 2 …..(ii)
2 cosec 2Q
From (i) and (ii)
p 2m P cos Q
F1 F1
KH ; 0
t t
p2 1
KH m 2 q 2 put the value of p and q
2m 2
K P using equation of motion Q and P
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K
P 0 P 0 P constant
Q
K K
Q Q Q constant P constant
P P
Q Q t from boundary condition 0
Q t
Therefore, option (d) is correct.
Q63. The Lagrangian of a particle moving in a plane s given in Cartesian coordinates as
L xy
x 2 y 2
In polar coordinates the expression for the canonical momentum pr (conjugate to the
radial coordinate r ) is
(a) r sin r cos (b) r cos r sin
(c) 2r cos r sin 2 (d) r sin 2 r cos 2
Ans.: (d)
Solution: L xy
x 2 y 2 xy
x 2 y 2
x r cos , y r sin x r cos r sin , y r sin r cos
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Q64. Let x, t and x, t be the coordinate systems used by the observers O and O ,
x x x x 1 1
(a) x and x (b) x x and x x
1 2
1 2 1 1
x x x x 1 1
(c) x and x (d) x x and x x
1 2 1 2 1 1
Ans: (d)
Solution: x x ct
vx v v v v v
ct 2 x 1 ct 1 1 1 1
x vt c c c
x c ct c c x ct
v2 v2 v 2
v 2 v v v
1 2 1 2 1 2 1 2 1 1 1
c c c c c c c
1
x x
1
vx v v
ct 2 x 1 ct 1
x vt c c c
x x ct
v2 v2 v2 v2
1 2 1 2 1 2 1 2
c c c c
v v
1 1
x x c ct c x 1 x ct x 1 x
1
v
1
v 1 1
c c
Q65. A ball of mass m , initially at rest, is dropped from a height of 5 meters. If the coefficient
of restitution is 0.9 , the speed of the ball just before it hits the floor the second time is
approximately (take g 9.8 m / s 2 )
(a) 9.80 m / s (b) 9.10 m / s (c) 8.91 m / s (d) 7.02 m / s
Ans: (c)
Solution: velocity just before hitting first time is
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v1 2 gh 2 9.8 5 9.89 m / s
v2 8.9 m / s
A 2 At
Putting this value of p in equation (iii) gives q e
B
Hence, the correct option is (a)
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Q67. A canonical transformation q, p Q, P is made through the generating function
F q, P q 2 P on the Hamiltonian
p2
H q, p q4
2 q 2
4
P 4P Q
(a) Q and P Q (b) Q and P
2
P 2P 2 2P
(c) Q and P Q (d) Q and P Q
Q
Ans.: (b)
Solution: F q, P q 2 P
F2 F
p & 2 Q
q P
1 1
p 2qP & Q q 2 q Q 2 & p 2 Q 2 P
4QP 2 2 2 P 2 Q 2
H Q, P Q
2 Q 4 4
H 4P H Q
Q Q and P P
P Q 2
Q68. The Lagrangian of a system moving in three dimensions is
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Solution: The motion is in 3D . So don’t get confine with x1 , x2 x3 they are actually x, y, z
Langrangian is then
L L L
mx m y 2 z 2 kx 2 k y z , when
1 2 1 1 2
L 0, 0, 0
2 2 2 x y z
So, not any component at Linear momentum is conserve.
Now transform the Lagrangian to Hamiltonian
Px2 Py
2
P2 1 1
z kx 2 k y z
2
H
2m 4m 4m 2 2
H
0 so energy is conserved
t
Now let us assume Lx yPz zPy
dLx L
Lx , H x
dt t
Lx , H yPz zPy , H y, H Pz y Pz , H z, H Py z Py , H
Py2 1 2 Pz2 1 2
Lx , H y, Pz y Pz , k y z z , Py z Py , k y z
4m 2 4m 2
Pz 1 P 1
2 Py y 0 k .2 y z 2 Py z z 0 k .2 y z
4m 2 4m 2
k y 2 yz k z 2 yz k y 2 z 2 k z 2 y 2
dLx dLy dL
0 . Similarly 0 and z 0
dt dt dt
Q69. For a particle of energy E and momentum p (in a frame F ), the rapidity y is defined
1 E p3c
as y ln . In a frame F moving with velocity v 0, 0, c with respect to
2 E p3c
F , the rapidity y will be
1 1
(a) y y ln 1 2
1
(b) y y ln
2 2 1
1 1
(c) y y ln (d) y y 2 ln
1 1
Ans: (b)
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1 E p3c
Solution: y ln
2 E p3c
1 E p3c
Then y ln
2 E p3c
E
Where p3 p3 v 2 E E vp3
c
v
1 E p3c E p3c
Put the value of p3 and E one will get y ln c
2 E p c v E p c
3 3
c
1 E p3 c 1 1 E p3 c 1 1
ln ln ln
2 E p3 c 1 2 E p3 c 2 1
1 1 1 1
y ln y ln
2 1 2 1
NET/JRF (DEC-2016)
Q70. A ball of mass m is dropped from a tall building with zero initial velocity. In addition to
gravity, the ball experiences a damping force of the form , where is its
instantaneous velocity and is a constant. Given the values m 10 kg , 10 kg / s and
(a) 10 t 1 e t (b) 10 t 1 e t
(c) 5t 2 1 et (d) 5t 2
Ans. : (b)
d2x dx
Solution: m mg
dt 2
dt
Putting the values of m, and g and simplifying we obtain
d 2 x dx
10
dt 2 dt
The general solution of this equation is x t c1 c2t 10e t
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We obtain, c1 10 and c2 10
Q71. A relativistic particle moves with a constant velocity v with respect to the laboratory
frame. In time , measured in the rest frame of the particle, the distance that it travels in
the laboratory frame is
c v2 v
(a) v (b) (c) v 1 (d)
v 2 c2 v2
1 2 1 2
c c
Ans. : (d)
Solution: From Particle x1 0 x2 0 tinitial t1 t final t2
(apart from the total energy of the particle) is also a constant of motion?
(a) p y 2 px (b) px 2 p y
(c) px 2 p y (d) p y 2 px
Ans. : (a)
Solution: V x, y x 2 y
px2 p y
2
H x 2y
2m 2m
d p y 2 px
dt
p y 2 px , H
t
p y 2 px
p y 2 px , H p y 2 px , x 2 y p y , 2 y 2 px , x 2 2 0
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Q73. The dynamics of a particle governed by the Lagrangian
1 2 1 2
L mx kx kxxt
describes
2 2
(a) an undamped simple harmonic oscillator
(b) a damped harmonic oscillator with a time varying damping factor
(c) an undamped harmonic oscillator with a time dependent frequency
(d) a free particle
Ans. : (d)
1 2 1 2
Solution: L mx kx kx x t
2 2
L L
mx kxt , kx kxt
x x
d L L
0 mx kxt 0 mx 0
kx kx kxt
dt x x
So motion is equivalent to free particle
Q74. The parabolic coordinates , are related to the Cartesian coordinates x, y by
x and y
1 2
2
2 . The Lagrangian of a two-dimensional simple harmonic
m 2 2 2 2
1 2
(a)
2
(b)
1
2
m 2 2 2 2 2 2 2
1
4
m 2 2 2 2 2
1 1
(c)
2 2
1
m 2 2 2 2 2
1
(d)
2 4
Ans. : (b)
Solution: For two dimensional Harmonic oscillation
m x 2 y 2 m 2 x 2 y 2
1 1
L
2 2
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y 2 2
1
x ,
2
,
x y
2 2 1 2 2 2
1
12 m
2 2
L m 2
2 4
L
1
2
m 2 2 2 2 2 2 2 2 m 2 4 4 2 2 2
1
8
1
m 2 2 2 2 m 2 2 2
1 2
2 8
m 2 2 2 2 2 2 2
1 1
2 4
Q75. After a perfectly elastic collision of two identical balls, one of which was initially at rest,
the velocities of both the balls are non zero. The angle between the final, velocities (in
the lab frame) is
(a) (b) (c) 0 (d)
2 2 2
Ans. : (a)
v1
Solution: Angle between two particle 1 2 0
Conservation of momentum u 1
2
mu mv1 cos 1 mv2 cos 2 (i)
v2
0 mv1 sin 1 mv2 sin 2 (ii)
conservation of kinetic energy
1 1 1
mu 2 mv12 mv22 (iii)
2 2 2
From (i) and (ii)
u 2 v12 v22 2v1v2 cos 1 cos 2 sin 1 sin 2
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cos 1 2 0
1 2
2 2
k
Q76. Consider circular orbits in a central force potential V r , where k 0 and
rn
0 n 2 . If the time period of a circular orbit of radius R is T1 and that of radius 2 R is
T2
T2 , then
T1
n 2 n
n 1
(a) 2 2 (b) 2 3 (c) 2 2 (d) 2n
Ans. : (c)
J2 k Veff J2 nk
Solution: Veff , n 1 0
2mr 2
r n
r mr 3
r
m 2 2 r 4 nk 1 n 2 / 2
n
1
J mr 2 n 1
2
n2
r T r 2
r 3
r r
n2
n
T2 2 R 2 1
2 2
T1 R
c
Q77. Consider a radioactive nucleus that is travelling at a speed with respect to the lab
2
frame. It emits -rays of frequency v0 in its rest frame. There is a stationary detector,
(which is not on the path of the nucleus) in the lab. If a -ray photon is emitted when the
nucleus is closest to the detector, its observed frequency at the detector is
3 1 1 2
(a) v0 (b) v0 (c) v0 (d) v0
2 3 2 3
Ans. : (a)
v2
Solution: v v0 1 (If detector is not in the path at nucleus)
c2
1 3
v v0 1 v0
4 2
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Q78. The Hamiltonian for a system described by the generalised coordinate x and generalised
momentum p is
p2 1
H x2 p 2 x2
2 1 2 x 2
1 2
x 2 x 1 2 x 2 x 2
1 1 1
2
(c) (d) x 2 2 x 2 x 2 x
2 2 2 1 2 x 2
Ans. : (a)
p2 1
Solution: H ax 2 p 2 x2 .
2 1 2 x 2
H p
x ax 2 p x ax 2 1 2 x
p 1 2 x
L xP
H
p2 1
xP
ax 2 P 2 x2
1 2 x 2
x x x
x x 1 2 x
2 2 2
2
1 2 x x x x 1 2 x
2 2
2 1 2 x
1 2 x x x 2 x x 2
x x2 1 2 x 2
2 2
1 2 x x x
x x 2 2
x x
2 2
2
2
1
2 x 2 1 2 x
2 2
1
2 x2
2
Q79. An inertial observer sees two events E1 and E2 happening at the same location but 6 s
apart in time. Another observer moving with a constant velocity v (with respect to the
first one) sees the same events to be 9 s apart. The spatial distance between the events,
as measured by the second observer, is approximately
(a) 300 m (b) 1000 m (c) 2000 m (d) 2700 m
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Ans. : (c)
Solution: x12 x11 0 , t21 t11 6 106 , t2 t1 9 106 , x2 x1 ?
t2 t2 9 106
1 v 1 t ' v x'
t2 c 2 x2 1 c 2 1
9 106
1 v / c 1 v / c
2 2 2 2
t2 t1 6 106
9 106 9 106
1 v / c
2 2
1 v / c
2 2
5 v2
v c 1 2 2 / 3
9 c
x2 vt2 x1 vt1
x2 x1
1 v / c 1 v / c
2 2 2 2
v
t2 t1
1 v2 / c2
c 6 106
5 9 5 9
x2 x1 3 108 6 106
3 6 3 6
9 5 102 20.12 102 2000m
Q80. A ball weighing 100 gm , released from a height of 5 m , bounces perfectly elastically off
a plate. The collision time between the ball and the plate is 0.5 s . The average force on
the plate is approximately
(a) 3 N (b) 2 N (c) 5 N (d) 4 N
Ans. : (d)
100
Solution: m 0.1 kg
1000
1 2
mgh mv v 2 gh .
2
v 10 m / sec
P 2
f 4N
t 0.5
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Q81. A solid vertical rod, of length L and cross-sectional area A , is made of a material of
Young’s modulus Y . The rod is loaded with a mass M , and, as a result, extends by a
small amount L in the equilibrium condition. The mass is then suddenly reduced to
M / 2 . As a result the rod will undergo longitudinal oscillation with an angular frequency
(a) 2YA / ML (b) YA / ML
YA
ml
Q82. The Lagrangian of a free relativistic particle (in one dimension) of mass m is given by
dP P
For constant force F , P Ft t
dt F
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mu
Ft u
F / m t .
1 u 2 / c2 Ft
2
1
mc
F
t. mc 2
2
dx m F t tdt Ft
x 1 1
dt Ft
2 m 0
Ft
2 F mc
1 1
mc mc
P2
2
Fx
2 1 1 2 4
mc mc
P 2 F 2 x 2 2mc 2 Fx Fx mc 2 m 2 c 4
2
Fx mc 2 2
P 2 m 2 c 4 , which is equation of hyperbola.
the time and is a constant, that is, the new coordinate is a combination of the old
coordinate and momentum at a shifted time. The new canonical momentum P t can be
expressed as
(a) p t q t (b) p t q t
1 1
(c) p t q t (d) p t q t
2 2
Ans. : (d)
Solution: Given q1 p 1
Q t q t p t
1 1 2
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1 2
For option (c) Q, P q p, p q 1 i.e. canonical transform
2 2
1 2
For option (d) Q, P q p, p q 1
2 2
Option (c) and (d) are correct. But from translation symmetry option (d) is more suitable.
Q84. The energy of a one-dimensional system, governed by the Lagrangian
1 2 1 2n
L mx kx
2 2
where k and n are two positive constants, is E0 . The time period of oscillation
satisfies
1 1 1 n 1 n2 1 1 n
(a) k n
(b) k 2n
E02 n (c) k 2n
E02 n (d) k n E02 n
Ans. : (b)
H J J
Solution: T , Time Period T , where J is Action variable
J H E
V x
J Pdx 4 2m E V x dx
0
1/ 2 n
x
2E
k
1
2E 2E
1/ 2 n
x
k
k
k 2n 2E 2E 2
1
J 4
0
2m E kx 2 n dn 4 2mE
2
0
1
2E
x dx k k
1 n
1 n
J 1 1. 1
T k E
2 n 2n
k E 2n
2 n
E
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ELECTROMAGNETIC THEORY
NET/JRF–(JUNE-2011)
Q1. The electrostatic potential V(x, y) in free space in a region where the charge density ρ is
zero is given by V x, y 4e 2 x f x 3 y 2 . Given that the x-component of the electric
(a) 3x 2 4e 2 x 8 x (b) 3x 2 4e 2 x 16 x
(c) 4e 2 x 8 (d) 3x 2 4e 2 x
Ans: (d)
Solution: V 4e 2 x f x 3 y 2 . Since 0 2V 0 16e 2 x f x 6 0 .
Since E x 0 at origin E V E x 8e 2 x f x
E x 0, 0 8 f 0 0 f 0 8 .
Since V 0, 0 0 4 f 0 0 f 0 4
f 0 8 c1 8 c1 0 .
Again Integrate f x 6 x 8e 2 x f x 3 x 2 4e 2 x c 2
since f 0 4 c2 4 c2 0 . Thus f x 3x 2 4e 2 x
Q2. For constant uniform electric and magnetic field E E 0 and B B0 , it is possible to
choose a gauge such that the scalar potential and vector potential A are given by
1 1
(a) 0 and A B0 r
2
(b) E 0 r and A B0 r
2
(c) E0 r and A 0 (d) 0 and A E 0 t
Ans: (a)
Solution: Let E E 0 xˆ yˆ zˆ and B B0 xˆ yˆ zˆ since they are constant vector.
Lorentz Gauge condition is A 0 0
t
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since B r B0 z y xˆ B0 z x yˆ B0 y x zˆ
(a) 0 and A 0 (b) 0, and A 0
t t
(c) 0 and A 0 (d) 0 and A 0
t t
Q3. A plane electromagnetic wave is propagating in a lossless dielectric. The electric field is
given by
E x, y, z, t E0 xˆ Azˆ exp ik0 ct x 3 z ,
where c is the speed of light in vacuum, E0 , A and k0 are constant and x̂ and ẑ are
unit vectors along the x - and z -axes. The relative dielectric constant of the medium r
and the constant A are
1 1
(a) r 4 and A (b) r 4 and A
3 3
k0c c
Since v Refractive index n r 2 r 4.
k k 3k
2
0
2
0
2
Since k nˆ 0 k 0 xˆ 3 zˆ xˆ Azˆ 0 k 0 1 A 3 0 A 1
3
A Kr
Q4. A static, spherically symmetric charge distribution is given by r e where A
r
and K are positive constants. The electrostatic potential corresponding to this charge
distribution varies with r as
(a) re Kr (b)
1 Kr
r
e (c)
1 Kr
r2
e (d)
1
r
1 e Kr
Ans: (b)
Solution: since 2V / 0
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A 1 2 V
2V must be proportional to e kr , where 2V 2 r .
r r r r
Q5. The magnetic field of the TE11 mode of a rectangular waveguide of dimensions a b as
shown in the figure is given by H z H 0 cos 0.3 x cos 0.4 y , where x and y are in
x
cm.
a
z
b
y
m n
0.3 where m 1 and 0.4 where n 1
a b
a 3.33cm, b 2.50cm
B. The entire range of frequencies f for which the TE11 mode will propagate is
(a) 6.0 GHz f 12.0 GHz (b) 7.5 GHz f 9.0 GHz
(c) 7.5 GHz f 12.0 GHz (d) 7.5GHz f
Ans: (d)
2 2
c m n c 1 1
Solution: f m , n f1,1 2 7.5 GH z .
2 a b 2 a 2
b
For propagation, frequency of incident wave must be greater than cutoff frequency.
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NET/JRF -(DEC-2011)
Q6. Consider three polarizer’s P1 , P2 and P3 placed along an axis as shown in the figure.
P1 P2 P3
(unpolarized)
I0
The pass axis of P1 and P3 are at right angles to each other while the pass axis of P2
I0 I0 I0
(a) 0 (b) (c) sin 2 2 (d) sin 2 2
2 8 4
Ans: (c)
Solution: I I 0 cos 2 (Malus Law)
I0 I0 I0 I
I1 , I2 cos 2 , I3 cos 2 cos 2 90 0 sin 2 2 .
2 2 2 8
Q7. Four equal point charges are kept fixed at the four vertices of a square. How many neutral
points (i.e. points where the electric field vanishes) will be found inside the square?
(a) 1 (b) 4 (c) 5 (d) 7
Ans: (a)
Solution: Inside the square, there is only one point where field vanishes.
Q8. A static charge distribution gives rise to an electric field of the form E 1 e r / R rˆ ,
r2
where and R are positive constants. The charge contained within a sphere of radius R ,
centred at the origin is
e e2 R R2
(a) 0 (b) 0 (c) 4 0 (d) 0
R2 R2 e e
Ans: None of the options given are correct
2
Solution: Qenc 0 E da 0 1 e r / R 2 r 2 sin ddrˆ 0 1 e r / R sin dd
rˆ
r 0 0
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1
at r R , Qenc 4 0 1 . So none of the options given are correct.
e
Q9. In a Young’s double slit interference experiment, the slits are at a distance 2 L from each
other and the screen is at a distance D from the slits. If a glass slab of refractive index
and thickness d is placed in the path of one of the beams, the minimum value of d for
the central fringe to be dark is
D D
(a) (b)
1 D 2 L2 1L
(c) (d)
1 2 1
Ans: (d)
n
Solution: For central fringe to be dark, 1d d
2 2 1
Q10. Consider a solenoid of radius R with n turns per unit length, in which a time dependent
current I I 0 sin t (where R / c 1 ) flows. The magnitude of the electric field at a
perpendicular distance r R from the axis of symmetry of the solenoid, is
1
(a) 0 (b) 0 nI 0 R 2 cos t
2r
1 1
(c) 0 nI 0 r sin t (d) 0 nI 0 r cos t
2 2
Ans: (d)
Solution: E d l
B
t d a ; B
0 nI t zˆ .
dI
r
2r 2
dt r 0
E 2r 0 n 2r d r 0 n I 0 cos t
2
1
E 0 nI 0 r cos t
2
Q11. A constant electric current I in an infinitely long straight wire is suddenly switched on at
t 0 . The vector potential at a perpendicular distance r from the wire is given
by A
kˆ 0 I 1
2
ln ct c 2 t 2 r 2
r
. The electric field at a distance r ct is
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I 1 ˆ ˆ
(a) 0 (b) 0
2 t 2
ij
c 0 I c 0 I
(c)
1 ˆ ˆ
i j (d) kˆ
2 c t r
2 2 2
2 2 c t r
2 2 2
Ans: (d)
A A I r 1 2c 2 t
c .
Solution: E
t
t
E 0
2 ct c 2 t 2 r 2 r
2 c 2t 2 r 2
c 0 I
E kˆ
2 c t r
2 2 2
NET/JRF -(JUNE-2012)
1 10
Q12. The magnetic field corresponding to the vector potential A F r 3 r , where F is
2 r
a constant vector, is
30 30
(a) F (b) F (c) F 4 r (d) F 4 r
r r
Ans: (a)
1
r
Solution: B A F r 10 3 . Since F is a constant vector, let
2 r
xˆ yˆ zˆ
F F0 xˆ yˆ zˆ , F r F0 F0 F0 xˆ z y F0 yˆ z x F0 zˆ y x F0
x y z
xˆ yˆ zˆ
F r
x
y
z
xˆF0 F0 yˆ F0 F0 zˆF0 F0 2 F0 xˆ yˆ zˆ
z y F0 x z F0
y x F0
r
1
F r F0 xˆ yˆ zˆ F , 3 0 . Thus B F
2 r
Q13. An electromagnetic wave is incident on a water-air interface. The phase of the
perpendicular component of the electric field, E , of the reflected wave into the water is
found to remain the same for all angles of incidence. The phase of the magnetic field H
(a) does not change (b) changes by 3 / 2
(c) changes by / 2 (d) changes by
Ans: (d)
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Q14. The magnetic field at a distance R from a long straight wire carrying a steady current I
is proportional to
(a) IR (b) I / R 2 (c) I 2 / R 2 (d) I / R
Ans: (d)
Q15. Which of the following questions is Lorentz invariant?
2 2 2 2 2 2 2
(a) E B (b) E B (c) E B (d) E B
Ans: (b)
Q16. Charges Q, Q and 2Q are placed on the vertices of an equilateral triangle ABC of
sides of length a , as shown in the figure. The dipole moment of this configuration of
charges, irrespective of the choice of origin, is - 2Q
ĵ C
(a) 2aQ iˆ a
a
(b) 3aQ ˆj
A B
Q a Q
(c) 3aQ ˆj
(d) 0 iˆ
Ans: (c)
Solution: Let coordinates of A is (l, m), then
a 3a ˆ
p qi ri Q liˆ mˆj Q l a iˆ mˆj 2Q l iˆ m j
2
2
p Q liˆ mˆj Q l a iˆ mˆj Q 2l a iˆ 2m 3a ˆj p 3aQˆj
mr
Q17. The vector potential A due to a magnetic moment m at a point r is given by A .
r3
If m is directed along the positive z -axis, the x - component of the magnetic field, at the
point r , is
3myz 3mxy 3mxz 3mz 2 xy
(a) (b) (c) (d)
r5 r5 r5 r5
Ans: (c)
Solution: m mzˆ and
m
B A 3 2 cos rˆ sin ˆ 3 3m rˆ rˆ m
r r
1
1
xxˆ yyˆ zzˆ r 3mxz
B 3 3mzˆ mzˆ Bx 5
r r r r
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NET/JRF -(DEC-2012)
Q18. Three charges are located on the circumference of a circle of radius R as shown in the
figure below. The two charges Q subtend an angle 900 at the centre Q Q
of the circle. The charge q is symmetrically placed with respect to
the charges Q . If the electric field at the centre of the circle is zero,
what is the magnitude of Q ?
q
(a) q / 2 (b) 2q (c) 2q (d) 4q
Ans: (a)
1 Q 1 q
Solution: E1 E 2 and E3
4 0 R 2
4 0 R 2
q
Resultant of E1 and E 2 is E E12 E 22 2E1 , Thus E3 E Q
2
Q19. Consider a hollow charged shell of inner radius a and outer radius b . The volume
k
charge density is r ( k is constant) in the region a r b . The magnitude of the
r2
electric field produced at distance r a is
k b a
(a) for all r a ,
0r 2
k b a kb
(b) for a r b and for r b
0r 2
0r 2
k r a k b a
(c) for a r b and for r b
0r 2
0r 2
k r a k b a
(d) for a r b and for r b
0a 2
0r 2
Ans: (c)
1 1 1 k
Solution: For r a : E.da E (4 r 2 ) Qenc dV 2 r sin drd d
2
0 0 0 r
4 k r 4 k k ra
E (4 r 2 ) a dr (r a) E rˆ
0 0 0 r 2
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4 k b 4 k k ba
For r b : E 4 r 2 a dr (b a) E rˆ
0 0 0 r 2
Q20. Consider the interference of two coherent electromagnetic waves whose electric field
vectors are given by E1 iˆE 0 cos t and E 2 ˆjE 0 cos t where is the phase
0
difference. The intensity of the resulting wave is given by E 2 , where E 2 is the
2
time average of E 2 . The total intensity is
(a) 0 (b) 0 E 02 (c) 0 E 02 sin 2 (d) 0 E 02 cos 2
Ans: (a)
Solution: Since waves are polarized in perpendicular direction hence there will be no
interference.
Q21. Four charges (two q and two q ) are kept fixed at the four vertices of a square of side
q q
Ans: (c)
Solution: Given configuration is quadrupole.
Q22. A point charges q of mass m is kept at a distance d below a grounded infinite
conducting sheet which lies in the xy - plane. For what value of d will the charge
remains stationary?
(a) q / 4 mg 0 (b) q / mg 0
Ans: (a)
Solution: There is attractive force between point charge q and grounded conducting sheet that
1q2 q
can be calculate from method of images i.e. mg d
4 0 2d 2
4 mg 0
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Q23. An infinite solenoid with its axis of symmetry along the z -direction carries a steady
current I .
ẑ
The vector potential A at a distance R from the axis
(a) is constant inside and varies as R outside the solenoid
R
(b) varies as R inside and is constant outside the solenoid
1
(c) varies as inside and as R outside the solenoid
R
1
(d) varies as R inside and as outside the solenoid
R
Ans: (d)
Q24. Consider an infinite conducting sheet in the xy -plane with a time dependent current
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NET/JRF -(JUNE-2013)
Q26. A particle of charge e and mass m is located at the midpoint of the line joining two fixed
collinear dipoles with unit charges as shown in the figure. (The particle is constrained to
move only along the line joining the dipoles). Assuming that the length of the dipoles is
much shorter than their separation, the natural frequency of oscillation of the particle is
R R
e, m
2d 2d
Ans: (c)
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Solution: Let charge per unit length be , hence I u in z -direction.
0 I ˆ
The magnetic field at a distance r is B .
2r
I
The electric field at a distance r is E rˆ rˆ .
2 0 r 2 0 ur
EB I2
Hence Poynting vector S zˆ
0 4 2 0 ur 2
Q28. If the electric and magnetic fields are unchanged when the potential A changes (in
suitable units) according to A A r̂ , where r r t r̂ , then the scalar potential must
simultaneously change to
(a) r (b) r (c) r / t (d) r / t
Ans: (c)
Solution: A A A rˆ / r 1 r C
V V / t V r / t
Q29. Consider an axially symmetric static charge distribution of the form,
2
r
0 0 e r / r0 cos 2
r
The radial component of the dipole moment due to this charge distribution is
(a) 2 0 r04 (b) 0 r04 (c) 0 r04 (d) 0 r04 / 2
Ans: (a)
2
r
Solution: p r r d r 0 0 e r / r0 cos 2 r 2 sin dr dd
V r
2
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Ans: (c)
Solution: A , Ax , Ay , Az kxyz, Ax kyzt , Ay kzxt , Az kxyt
A
Since k yzxˆ xzyˆ xyzˆ and k yzxˆ xzyˆ xyzˆ
t
A
E k yzxˆ xzyˆ xyzˆ k yzxˆ xzyˆ xyzˆ 0 E 0,0,0
t
Q31. An oscillating current I t I 0 exp it flows in the direction of the y -axis through a
thin metal sheet of area 1.0 cm 2 kept in the xy -plane. The rate of total energy radiated
per unit area from the surfaces of the metal sheet at a distance of 100 m is
Ans: (d)
NET/JRF -(DEC-2013)
Q32. A horizontal metal disc rotates about the vertical axis in a uniform magnetic field
pointing up as shown in the figure. A circuit is made by connecting one end A of a
resistor to the centre of the disc and the other end B to its edge through a sliding contact.
The current that flows through the resistor is
B
A B
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Q34. The electric field of an electromagnetic wave is given by
E E 0 cos 0.3x 0.4 y 1000t kˆ .
The associated magnetic field B is
(a) 10 3 E 0 cos 0.3 x 0.4 y 1000t kˆ
(b) 10 4 E 0 cos 0.3 x 0.4 y 1000t 4iˆ 3 ˆj
(c) E cos 0.3 x 0.4 y 1000t 0.3iˆ 0.4 ˆj
0
Ans: (b)
Solution: k 0.3xˆ 0.4 yˆ , 1000
kE
0.3 xˆ 0.4 yˆ E 0 cos 0.3x 0.4 y 1000t kˆ
1
B
B 10 4 E 0 cos 0.3x 0.4 y 1000t 4iˆ 3 ˆj
Q35. A point charge q is placed symmetrically at a distance d from two perpendicularly
placed grounded conducting infinite plates as shown in the figure. The net force on the
charge (in units of 1 / 4 0 ) is
q
(a)
q2
8d 2
2 2 1 away from the corner
d
d
(b)
q2
8d 2
2 2 1 towards the corner
q2
(c) towards the corner
2 2d 2
3q 2 F3
(d) away from the corner d d
8d 2 q q
Ans: (b) F1
F2 d
q2 q2
Solution: F 1 F 2 k and F 3 k
4d 2 8d 2 d
q 2 q q
Resultant of F 1 , F 2 is F12 F12 F22 2 2k . 2d
8d 2
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q
2
Net force F k 2 2 2 1 (towards the corner)
8d
Q36. If the electrostatic potential V r , , in a charge free region has the form
V r , , f r cos , then the functional form of f r (in the following a and b are
constants) is:
b b b r
(a) ar 2 (b) ar (c) ar (d) a ln
r r2 r b
Ans: (b)
1 V 1 V 1 2V
Solution: V 2 r 2 2 sin 2 2 2 0
2
r r r r sin r sin
1 2 f 1
r cos 2 sin f sin 0
r r r r sin
2
cos 2 2 f f f
r 2 2r 2 2sin cos 0
r2 r r r sin
2 f f
r2 2r 2 f r 0
r
2
r
b
f r ar satisfy the above equation.
r2
Q37. Let four point charges q, q / 2, q and q / 2 be placed at the vertices of a square of
side a . Let another point charge q be placed at the centre of the square (see the figure).
q/2 q
q
q q/2
Let V r be the electrostatic potential at a point P at a distance r a from the centre
of the square. Then V 2r / V r is
1 1 1
(a) 1 (b) (c) (d)
2 4 8
Ans: (d)
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q q
Solution: According to multipole expansion Qmono q q q 0
2 2
q q q q q q q qq q
p q xˆ yˆ xˆ yˆ 0 q xˆ yˆ xˆ yˆ 0
2 2 2 2 2 2 2 22 2
1 V 2r 1
Thus, V .
r 3
V r 8
Q38.
Let V , A and V , A denote two sets of scalar and vector potentials, and is a scalar
function. Which of the following transformations leave the electric and magnetic fields
(and hence Maxwell’s equations) unchanged?
(a) A A and V V (b) A A and V V 2
t t
(c) A A and V V (d) A A and V V
t t
Ans: (a)
JRF/NET–(JUNE-2014)
Q39. A time-dependent current I t Ktzˆ (where K is a constant) is switched on at t 0 in
an infinite current-carrying wire. The magnetic vector potential at a perpendicular
distance a from the wire is given (for time t a / c ) by
0 K c t a K
2 2 2
ct
ct a 2 z 2 t
4 c c t a dz a
(a) zˆ dz (b) zˆ 0
2 2 a 2 z 2 1 / 2
2
4 ct
2
z2 1/ 2
c 2t 2 a 2
K ct ct a 2 z 2 K t
(c) zˆ 0 dz (d) zˆ 0 dz
4 c ct a 2 z 2 1 / 2 4 c 2t 2 a 2 a 2
z2
1/ 2
Ans: (a) I
dz
I tr 0
K t R / c
Solution: A zˆ 0
4
R
dz zˆ
4
R
dz z R
P
0 K c t a
2 2 2
ct a 2 z 2 a
4 c c2t 2 a2
A zˆ dz
1/2
a2 z2
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Q40. A current i p flows through the primary coil of a transformer. The graph of i p t as a
1 2 3 t
Which of the following graphs represents the current i S in the secondary coil?
is t is t
(a) (b)
1 2 3 t
1 2 3 t
is t is t
(c) (d)
1 2 3t
1 2 3t
Ans: (c)
di p
Solution: is
dt
Q41. If the electrostatic potential in spherical polar coordinates is
r 0 e r / r
0
where 0 and r0 are constants, then the charge density at a distance r r0 will be
0 0 e 0 0 0 0 2e 0 0
(a) (b) (c) (d)
er 0
2
2r02 er 0
2
r02
Ans: (a)
Solution: 2 0 2
0
1 2 1 2 0 r / r0 1 0 2 r / r0
2 2 r 2 r e
r r r r r r0
2
r r r
r e
0
1 0 2 1 r / r0 1 2
2 r e 2re r / r0 0 e r / r0 e r / r0
r r0
2
r0 r0 r0 r
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1 2
At a distance r r0 , 2 0 e 1 e 1 20 0 20 02 0
r0 r0 r0 r0 e r0 e r0 e
Q42. If A yziˆ zxˆj xykˆ and C is the circle of unit radius in the plane defined by z 1 ,
with the centre on the z - axis, then the value of the integral A d is
C
(a) (b) (c) (d) 0
2 4
Ans: (d)
iˆ ˆj kˆ
Solution: A iˆ x x ˆj y y kˆ z z 0
x y z
yz zx xy
Since A d A d a 0
C
S
Q43. Consider an electromagnetic wave at the interface between two homogenous dielectric
media of dielectric constants 1 and 2 . Assuming 2 1 and no charges on the surface,
the electric field vector E and the displacement vector D in the two media satisfy the
following inequalities
(a) E 2 E1 and D2 D1 (b) E 2 E1 and D2 D1
(c) E 2 E1 and D2 D1 (d) E 2 E1 and D2 D1
Ans. : (c)
Q44. A charge e is placed in vacuum at the point d ,0,0 , where d 0 . The region x 0
d
is filled uniformly with a metal. The electric field at the point ,0,0 is
2
10e 10e
(a) 1, 0, 0 (b) 1, 0, 0
9 0 d 2
9 0 d 2
e e
(c) 1, 0, 0 (d) 1, 0, 0
0 d 2 0 d 2
Ans: (b)
Solution: E
d
E
2 P
x
e d 0 d e
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1 e 1 4e 1 e 1 4e
E and E
4 0 3d / 2 2
4 0 9d 2
4 0 d / 2 2
4 0 d 2
(a) Ax , Ay , Az
y x
, ,0
2 x y 2 x y
2 2 2 2
(b) Ax , Ay , Az
y x
, ,0
2 x y z 2 x y z
2 2 2 2 2 2
x y x y
(c) Ax , Ay , Az , ,0
2 x y 2 x y
2 2 2 2
(d) Ax , Ay , Az
x y
, ,0
2 x y 2 x y
2 2 2 2
Ans: (a)
Solution: B A 0
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Q47. An electromagnetically-shielded room is designed so that at a frequency 10 7 rad/s
the intensity of the external radiation that penetrates the room is 1% of the incident
I0 1 1
where 100, 106 4 107 107 103
I 2 2 2
1
z ln 100 2.30 mm
2 103
Q48. A charged particle is at a distance d from an infinite conducting plane maintained at zero
potential. When released from rest, the particle reaches a speed u at a distance d / 2 from
the plane. At what distance from the plane will the particle reach the speed 2u ?
(a) d / 6 (b) d / 3 (c) d / 4 (d) d / 5
Ans: (d) x
d x
2
1 q d x2
A2
q 2
Solution: F ma m 2 2 where A . P q
dt 2
4 0 4d 2
dt x 16 m 0
d
dv A dv A dx 1 d 2 d A
dt
2 v
x dt
2
x dt
2 dt
v
dt x
0
v2 A A 1 1 d
C at x d , v 0 C v 2 A .
2 x d x d q
1 1 2A 1 1 d
Thus u 2 A then 2u 2 A x
d /2 d d x d 5
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NET/JRF–(DEC-2014)
Q49. A charged particle moves in a helical path under the influence of a constant
magnetic field. The initial velocity is such that the component along the
magnetic field is twice the component in the plane normal to the magnetic l
field.
The ratio / R of the pitch to the radius R of the helical path is 2R
(a) / 2 (b) 4 (c) 2 (d)
Ans: (b)
Solution: v 2v
2 R 2 R l
Pitch of the helix l vT v 2v 4 R 4
v v R
Q50. A parallel beam of light of wavelength is incident normally on a thin polymer film
with air on both sides. If the film has a refractive index n 1 , then second-order bright
fringes can be observed in reflection when the thickness of the film is
(a) / 4n (b) / 2n (c) 3 / 4n (d) / n
Ans: (c)
Solution: For constructive interference: 2nd cos 2m 1
2
For normal incidence 0 and second order m 1
3
2nd cos 0 2 1 1 d
2 4n
Q51. A solid sphere of radius R has a charge density, given by
ar
r 0 1
R
where r is the radial coordinate and 0 , a and R are positive constants. If the
magnitude of the electric field at r R / 2 is 1.25 times that at r R , then the value of a
is
(a) 2 (b) 1 (c) 1 / 2 (d) 1 / 4
Ans: (b)
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r
1 1 ar
Solution: E .d a Qenc E 4 r 2 0 1 4 r 2 dr
S
0 0 0 R
40 2 ar 3 4 0 r 3 ar 4 0 r ar 2
r
0 0
E 4 r 2
r dr E
R 0 3 4R 0 3 4R
0 R / 2 aR 2 / 4 0 R aR 2
Er R / 2 1.25 Er R 1.25
0 3 4R 0 3 4R
1 a 5 1 a 1 a 5 5a 5a a 5 1
6 16 4 3 4 6 16 12 16 16 16 12 6
4a 5 2 a 3
a 1
16 12 4 12
Q52. The electrostatic lines of force due to a system of four point charges
is sketched here. At large distance r , the leading asymptotic
behaviour of the electrostatic potential is proportional to
(a) r (b) r 1
(c) r 2 (d) r 3
Ans: (d)
1
Solution: The given electrostatic line of force is due to a quadrupole. So V .
r3
Q53. A plane electromagnetic wave incident normally on the surface of a material is partially
reflected. Measurements on the standing wave in the region in front of the interface such
that the ratio of the electric field amplitude at the maxima and the minima is 5. The ratio
of the reflected intensity to the incident intensity is
(a) 4 / 9 (b) 2 / 3 (c) 2 / 5 (d) 1 / 5
Ans: (a)
E0 I E0 R E 2
Solution: 5 E0 I E0 R 5 E0 I E0 R 6 E0 R 4 E0 I 0 R
E0 I E0 R E0 I 3
2
I E 4
R 0R
I I E0 I 9
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Q54. A non-relativistic particle of mass m and charge e , moving with a velocity v and
acceleration a , emits radiation of intensity I . What is the intensity of the radiation
emitted by a particle of mass m / 2 , charge 2e , velocity v / 2 and acceleration 2a ?
(a) 16 I (b) 8 I (c) 4 I (d) 2 I
Ans: (a)
q 2 a 2 sin 2 I 2 q22 a22 I 2 4e 2 4a 2
Solution: I 16 I 2 16 I
r2 I1 q12 a12 I e2 a 2
NET/JRF–(JUNE-2015)
Q55. A Plane electromagnetic wave is travelling along the positive z -direction. The maximum
electric field along the x - direction is 10 V / m . The approximate maximum values of the
power per unit area and the magnetic induction B , respectively, are
(a) 3.3 107 watts / m 2 and 10 tesla
(b) 3.3 107 watts / m 2 and 3.3 108 tesla
(c) 0.265 watts / m 2 and 10 tesla
Q56.
Which of the following transformations V , A V ', A ' of the electrostatic potential
V and the vector potential A is a gauge transformation?
(a) V V ax, A A at kˆ
(b) V V ax, A A at kˆ
(c) V V ax, A A at iˆ
(d) V V ax, A A at iˆ
Ans. (d)
Solution: V V ax axt c
t t
atiˆ 0 . Thus, A A atxˆ
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Q57. Suppose the yz -plane forms a chargeless boundary between two media of permittivities
left and right where left :right 1: 2 , if the uniform electric field on the left is
Eleft c iˆ ˆj kˆ (where c is a constant), then the electric field on the right Eright is
(a) c 2iˆ ˆj kˆ
(b) c iˆ 2 ˆj 2kˆ
1 1 1
(c) c iˆ ˆj kˆ (d) c iˆ ˆj kˆ
2 2 2
Ans. (c) y
1 2
Solution: E1 c ˆj kˆ E2
1
D1 D2 1 E1 2 E2 E21 E1
2 x
1 ˆ 1
E2 ci E2 c iˆ ˆj kˆ
2 2 z
Q58. A proton moves with a speed of 300 m / s in a circular orbit in the xy -plan in a magnetic
field 1 tesla along the positive z - direction. When an electric field of 1 V / m is applied
along the positive y -direction, the center of the circular orbit
(a) remains stationary
(b) moves at 1 m / s along the negative x direction
(c) moves at 1 m / s along the positive z direction
(d) moves at 1 m / s along the positive x direction
z
Ans. (d)
Solution: change particle will deflect in x -direction with B
E 1 y
v 1 m/ s .
B 1 E
x
Q59. Consider a rectangular wave guide with transverse dimensions 2 m 1 m driven with an
angular frequency 109 rad / s . Which transverse electric TE modes will propagate
(c) TE01 , TE10 and TE11 (d) TE01 , TE10 and TE22
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Ans. (a)
m2 n2
Solution: mn C
a 2 b2
c 3 108 3.14
10 4.71 108 rod / sec
a 2
c 3 108 3.14
01 9.42 108 rod / sec
b 1
1 1
11 c 2 10.53 108 rod / sec
a b
2
2c
20 9.72 108 rod / sec
a
4 4
22 c 2 10.5 108 rod / sec
a b
2
Q60. The electric and magnetic fields in the charge free region z 0 are given by
E r , t E0 e k1z cos k2 x t ˆj
E
B r , t 0 e k1z k1 sin k2 x t iˆ k2 cos k2 x t kˆ
where , k1 and k2 are positive constants. The average energy flow in the x -direction is
E02 k2 2 k1z E02 k2 E02 k1 2 k1z 1
(a) e (b) e 2 k1z (c) e (d) c 0 E02 e 2 k1 z
20 0 20 2
Ans. (a)
1 E 2 e 2 k1 z
Solution: S
0
EB 0
0 1
k cos sin kˆ k2 cos 2 iˆ , where k2 x t
k E 2 e 2 k1 z E02 k2 2 k1 z
S 2 0 e
2 0 20
Q61. A uniform magnetic field in the positive z -direction passes through a circular wire loop
of radius 1 cm and resistance 1 lying in the xy -plane. The field strength is reduced
from 10 tesla to 9 tesla in 1s . The charge transferred across any point in the wire is
approximately
(a) 3.1104 coulomb (b) 3.4 104 coulomb
(c) 4.2 104 coulomb (d) 5.2 104 coulomb
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Ans. (a)
d dq 1 d A r 2
Solution: I dq dB dB
dt dt R R dt R R
2
3.14 102
dq 1 3.14 104 coulomb
1
Q62. A rod of length L carries a total charge Q distributed uniformly. If this is observed in a
frame moving with a speed v along the rod, the charge per unit length (as measured by
the moving observer) is
Q v2 Q v2 Q Q
(a) 1 2 (b) 1 2 (c) (d)
L c L c v2 v2
L 1 L 1 2
c2 c
Ans. (c)
0 Q
Solution:
v2
v2
1 L 1
c2 c2
NET/JRF–(DEC-2015)
Q63. A hollow metallic sphere of radius a , which is kept at a potential V0 has a charge Q at its
centre. The potential at a point outside the sphere, at a distance r from the centre, is
Q Va Q V a2 V0 a
(a) V0 (b) 0 (c) 02 (d)
4 0 r r 4 0 r r r
Ans.: (d)
Q
Solution: Let charge on conductor is Q , then V0
4 0 a
Q Va
Now V V 0
4 0 r r
Q64. Consider a charge Q at the origin of 3 - dimensional coordinate system. The
flux of the electric field through the curved surface of a cone that has a height
h
h and a circular base of radius R (as shown in the figure) is
R Q
Q Q hQ QR
(a) (b) (c) (d)
0 2 0 R 0 2h 0
Ans.: (b)
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Q65. Given a uniform magnetic field B B0 kˆ (where B0 is a constant), a possible choice for
the magnetic vector potential A is
plane interface formed with another medium of dielectric constant 2 3 1 . The two
media have identical magnetic permeability. If the angle of incidence is 600 , then the
reflected light
(a) is plane polarized perpendicular to the plane of incidence
(b) is plane polarized parallel to the plane of incidence
(c) is circularly polarized
(d) has the same polarization as the incident light
Ans.: (a)
n I
Solution: B tan 1 2
1
n1
2
2
B tan 1
tan
1
3
1
B 600 (hence reflected light is plane polarized perpendicular to plane of incidence))
Q67. A small magnetic needle is kept at 0, 0 with its moment along the x -axis. Another
small magnetic needle is at the point 1,1 and is free to rotate in the xy - plane. In
following is true? (Assume that 1 ).
0
(a) There is negligible absorption of the wave
(b) The wave propagation is highly dispersive
(c) There is strong absorption of the electromagnetic wave
(d) The group velocity and the phase velocity will have opposite sign
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Ans.: (a)
Solution: When 0 , there is negligible absorption of the wave.
NET/JRF–(JUNE-2016)
Q70. Four equal charges of Q , each are kept at the vertices of a square of side R . A particle
of mass m and charge Q is placed in the plane of the square at a short distance
a R from the centre. If the motion of the particle is confined to the plane, it will
Q2 Q2
(a) (b)
2 0 R 3 m 0 R 3m
2Q 2 Q2
(c) (d)
0 R 3 m 4 0 R 3m
Ans: (c)
kQ Q Q
Solution: E1 E2
R R2
2
E2 E1 y
a
2 4
a x
Resultant field E12, y 2 E1 cos
R /2
2kQ R 2kQ R
E12, y a a Q R /2 Q
2 2
3 3
R R2
2 2 R2 2
a
2 4 2
4 2kQ R
E12, y a
R 3
2
Q R /2 Q
kQ R
Similarly; E3 E4 2 a
R R
2 2
a
2 4 E3 a E4
2kQ R
Resultant E34, y 2 E3 cos a
2
3
R R
2 2 2
a Q Q
2 4
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4 2kQ R
E34, y a
R 3
2
4 2kQ R R 8 2kQ
Resultant E a a a
R 3
2 2 R3
8 2 1 2 2Q
E Qa E a
R 3
4 0 0 R 3
2 2Q 2 2 2Q 2
F QE a
0 R 3 0 mR 3
Q71. Two parallel plate capacitors, separated by distances x and 1.1x respectively, have a
dielectric material of dielectric constant 3.0 inserted between the plates and are
connected to a battery of voltage V . The difference in charge on the second capacitor
compared to the first is
(a) 66% (b) 20% (c) 3.3% (d) 10%
Ans: (d)
3 0 A 3 A
Solution: Q1 C1V1 V , Q2 C2V2 0 V
x 1.1x
1 3 A
1 0 V
Q2 Q1 x
100%
1.1
100 9%
Q1 3 A
0
V
x
Q72. The half space region x 0 and x 0 are filled with dielectric
x0 x0
media of dielectric constants 1 and 2 respectively. There is a 2 1
uniform electric field in each part. In the right half, the electric field 1 E1
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E1
tan 1 E E
Solution:
tan 2
1 1
E E2
E1 E2
2
E2
E1 2 tan 1 2
D1 D2 1 E1 2 E2 1 tan 1 2 tan 2
E2 1
tan 2 1
1
(c) By B0 x 2 y 2 (d) By B0 x3 xy 2
3
Ans: (b)
Solution: Bx B0 x 2 y 2 , Bz 0
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Q75. Consider a sphere S1 of radius R which carries a uniform charge
S2
R S
of density . A smaller sphere S2 of radius a is cut out and 1 P
2
b
removed from it. The centres of the two spheres are separated by r
nR ˆ
the vector b , as shown in the figure. The electric field at a
2
point P inside S2 is
R R R a
(a) nˆ (b) r na
ˆ (c) nˆ (d) r
3 0 3 0 a 6 0 3 0 R
Ans: (c)
Solution: Electric field at P due to S1 is E1 r
3 0 S2
S1
Electric field at P due to S2 (assume ) is E2 r r
3 0
b P
Thus E E1 E2 r r ; b r r r r b r
3 0
R R
E b nˆ b nˆ
3 0 6 0 2
Q76. The value of the electric and magnetic fields in a particular reference frame (in Gaussian
units) are E 3xˆ 4 yˆ and B 3 zˆ respectively. An inertial observer moving with respect
to this frame measures the magnitude of the electric field to be E 4 . The magnitude of
Q77. A loop of radius a , carrying a current I , is placed in a uniform magnetic field B . If the
normal to the loop is denoted by n̂ , the force F and the torque T on the loop are
(a) F 0 and T a 2 I n̂ B (b) F 0 I B
4
1
(c) F 0 I B and T I nˆ B (d) F 0 and T IB
4 0 0
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Ans: (a)
Solution: In uniform field F 0
Torque T m B a 2 Inˆ B
Q78. A waveguide has a square cross-section of side 2a . For the TM modes of wave vector k ,
the transverse electromagnetic modes are obtained in terms of a function x, y which
mode is given by
4 2 2
(a) 2 c 2 k 2 2 (b) 2 c 2 k 2 2
a a
2 2 2 2
(c) c k 2
2 2
(d) c k 2
2 2
2a 4a
Ans: (c)
Solution: c 2 k 2 2 mn
2
2 c 2 k 2 mn
2
m2 n2 1 1
mn
2
c 2 2 2 2 112 c 2 2
a b 2a 2a
2 2
1 c 2 2 2 2 2
112 c 2 2 2
c k
2a 2 2a 2 2a 2
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NET/JRF -(DEC-2016)
Q79. A screen has two slits, each of width w with their centres at a distance 2 w apart. It is
illuminated by a monochromatic plane wave travelling along the x -axis.
The intensity of the interference pattern, measured on a distant screen, at an angle
n
to the x -axis is
w
w
(a) zero for n 1, 2,3...
w x
(b) maximum for n 1, 2,3...
1 3 5 w
(c) maximum for n , , ...
2 2 2
(d) zero for n 0 only
Ans. : (a)
Solution: maximum for n 0 and zero for n 1, 2,3... .
Q80. The electric field of an electromagnetic wave is
E z , t E0 cos kz t iˆ 2 E0 sin kz t ˆj
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Q81. A conducting circular disc of radius r and resistivity rotates with an angular velocity
in a magnetic field B perpendicular to it. A voltmeter is connected as shown in the
figure below. Assuming its internal resistance to be infinite, the reading on the voltmeter
(a) depends on , B, r and
B
(b) depends on , B and r but not on
(c) is zero because the flux through the loop is not
changing
r V
(d) is zero because a current the flows in the direction
of B
Ans. : (b)
Solution: Force experienced by charge is
F q v B and v r
Q82. The charge per unit length of a circular wire of radius a in the xy -plane, with its centre at
the origin, is 0 cos , where 0 is a constant and the angle is measured from the
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Q83. A pair of parallel glass plates separated by a distance d is illuminated by white light as
shown in the figure below. Also shown in the graph of the intensity of the reflected light
I as a function of the wavelength recorded by a spectrometer.
1
spectrometer
0.8
Intensity
partially incident 0.6
reflecting mirror white light 0.4
0.2
0
d air gap 490 500 510 520 530
glass plates m
Assuming that the interference takes place only between light reflected by the bottom
surface of the top plate and the top surface of bottom plate, the distance d is closest to
(a) 12 m (b) 24 m (c) 60 m (d) 120 m
Ans. : (d)
1
Solution: For constructive interference of reflected light, 2d cos n .
2
495 m
First maxima occurs at 495 m , 00 and n 0 . Thus, d 120 m
4 4
Q84. Suppose that free charges are present in a material of dielectric constant 10 and
resistivity 1011 m . Using Ohm’s law and the equation of continuity for charge, the
1
time required for the charge density inside the material to decay by is closest to
e
(a) 106 S (b) 106 S (c) 1012 S (d) 10 S
Ans. : (d)
t
8.8 1012 10
0r 1
Solution: f t f 0 e ;
, 10sec ,
10 11
Q85. A particle with charge q moves with a uniform angular velocity in a circular orbit of
radius a in the xy - plane, around a fixed charge q , which is at the centre of the orbit at
I2
The ratio for R a , is
I1
1 1
(a) 4 (b) (c) (d) 8
4 8
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Ans. : (c)
I 2 r13 R3 1
Solution: 3
I1 r2 2 R 3
8
Q86. A parallel plate capacitor is formed by two circular conducting plates of radius a
separated by a distance d , where d a . It is being slowly charged by a current that is
nearly constant. At an instant when the current is I , the magnetic induction between the
a
plates at a distance from the centre of the plate, is
2
0 I 0 I 0 I 0 I
(a) (b) (c) (d)
a 2 a a 4 a
Ans. : (d)
Ir P
Solution: B 0 2 I a
2 a r
I a
B 0 at r
4 a 2
Q87. Two uniformly charged insulating solid spheres A and B , both of radius a , carry total
charges Q and Q , respectively. The spheres are placed touching each other as shown
in the figure.
A B
If the potential at the centre of the sphere A is VA and that at the
centre of B is VB then the difference VA VB is
Q Q Q Q
(a) (b) (c) (d)
4 0 a 2 0 a 2 0 a 4 0 a
Ans. : (c)
3Q Q Q
Solution: VA
8 0 a 4 0 2a 4 0 a
3Q Q Q
VB
8 0 a 4 0 2a 4 0 a
Q
VA VB
2 0 a
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NET/JRF -(JUNE -2017)
Q88. Two long hollow co-axial conducting cylinders of radii R1 and R2 R1 R2 are placed
2 R1 2 R2
The inner cylinder carries a charge per unit length and the outer cylinder carries a
charge per unit length. The electrostatic energy per unit length of this system is
2 2
(a)
0
ln R2 / R1 (b)
4 0
R22 / R12
2 2
(c) ln R2 / R1 (d) ln R2 / R1
4 0 2 0
Ans. : (c)
Solution: r R1 , E1 0 ; R1 r R2 , E2 rˆ
2 0 r
r Rz , E3 0
0 Rz 2
W E 2 dz 0
2 all spce 2 R1 4 2 02 r 2
2 rldr
W 0 2 R2 1 2 R
l
2 2 02 R1 r
dr ln 2
4 0 R1
Q89. A set of N concentric circular loops of wire, each carrying a steady current I in the
same direction, is arranged in a plane. The radius of the first loop is r1 a and the radius
of the n th loop is given by rn nrn 1 . The magnitude B of the magnetic field at the centre
(c) 0 I e 2 1 / 8a (d) 0 I e 1 / 2a
Ans. : (d)
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I1 1 1 1
Solution: B 0 ........
2 r1 r2 r3 rn
r1 a
rn nrn 1
r1 r0 a
r2 2r1 2a
r3 3r2 3.2a
r4 4r3 4.3.2a
0 I 1 1 1
B 1 .......
2a 2 3.2 4.3.2
0 I 1 N
B
2a n 1 n
xn
1
1
1
ex e 1 e 1
n 0 n n 0 n n 1 n n 1 n
N 1 I
lim e 1 B 0 e 1
N
n l n 2a
Q90. An electromagnetic wave (of wavelength 0 in free space) travels through an absorbing
I
medium with dielectric permittivity given by R i I where 3 . If the skin
R
0
depth is , the ratio of the amplitude of electric field E to that of the magnetic field B ,
4
in the medium (in ohms) is
(a) 120 (b) 377 (c) 30 2 (d) 30
Ans. : (d)
1 0 I
Solution: d , 3
4 R
1/ 2
2
1
1
2
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4 2 4
2 0 0
1/ 2
2
K k 1
2 2
E0 1 1
B0 K 2 4
1/ 2
2
2
2
1 0
0 0 2 c / 0 c E c
0
8 8 4 H0 4
E 4 107 3 108
30
H0 4
Q91. The vector potential A ke at rrˆ (where a and k are constants) corresponding to an
electromagnetic field is changed to A ke at rrˆ . This will be a gauge transformation if
the corresponding change in the scalar potential is
0 e2 au
E
12 c
E 2E
Fraction of initial K .E. lost due to radiation
1
mu 2 mu 2
2
2 0e 2 au 0 e2 a
mu 2 12 c 6 mcu
1 2 u 1 u2 u 2 u 2 3u 2 3u 2
s ut at u a a
2 2a 2 4 a 2 2 a 8a 8a 8s
0 e2 3u 2 e 2u
0
6 mcu 8s 16 mcs
Q93. The figure below describes the arrangement of slits and screens in a Young’s double slit
experiment. The width of the slit in S1 is a and the slits in S2 are of negligible width.
S1 S2
screen
b
d
If the wavelength of the light is , the value of d for which the screen would be dark is
2 2 2
a b a ab ab
(a) b 1 (b) 1 (c) (d)
2 2
Ans. : (d)
Solution: If the path difference Op2 Op1
2
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The minima of the interference pattern produced by O will fall on the maxima produced
1/ 2
2 b a 2 1 b a
2
ab
OP2 OP1 d b, a P2
2d
ab ab
Thus d d
2 2d
Q94. A constant current I is flowing in a piece of wire that is bent into a loop as shown in the
figure. y
2b
2b a a
2b
2a
2a O x
b b
The magnitude of the magnetic field at the point O is
0 I a 0 I 1 1
(a) ln (b)
4 5 b 4 5 a b
0 I 1 0 I 1
(c) (d)
4 5 a 4 5 b
Ans. : (b)
I
Solution: B 0 sin 2 sin 1 ˆ
4 d
Magnetic field due to left and right segment of 2a
d 1 2
I 2a
B2 a 0
4 a 5a I
Field due to upper segment of 2a
0 I a a
4 2a 5a 5a
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I 2 0 I 1
Net field B2 a 2 0
4 a 5 4 a 5
0 I
B2 a 5 (inward)
4 a
0 I
similarly, B2b 5 (outward)
4 b
0 I 1 1
Net field B B2 a B2b 5
4 a b
Q95. The charge distribution inside a material of conductivity and permittivity at initial
time t 0 is r , 0 0 , a constant. At subsequent times r , t is given by
t 1 t
(a) 0 exp (b) 0 1 exp
2
0 t
(c) (d) 0 cosh
t
1 exp
Ans. : (a)
Solution: J f E , .E f , .J f f
E
.E f f f
t t
f t 0 exp f f t 0 exp t
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QUANTUM MECHANICS SOLUTIONS
NET/JRF (JUNE-2011)
1
Q1. The wavefunction of a particle is given by 0 i1 where 0 and 1 are the
2
normalized eigenfunctions with energies E0 and E1 corresponding to the ground state
and first excited state, respectively. The expectation value of the Hamiltonian in the state
is
E0 E0 E0 2 E1 E0 2 E1
(a) E1 (b) E1 (c) (d)
2 2 3 3
Ans: (d)
1 H E 0 2 E1
Solution: 0 i1 and H
2 3
Q2. The energy levels of the non-relativistic electron in a hydrogen atom (i.e. in a Coulomb
potential V r 1 / r ) are given by E nlm 1 / n 2 , where n is the principal quantum
number, and the corresponding wave functions are given by nlm , where l is the orbital
angular momentum quantum number and m is the magnetic quantum number. The spin
of the electron is not considered. Which of the following is a correct statement?
(a) There are exactly 2l 1 different wave functions nlm , for each Enlm .
(b) There are l l 1 different wave functions nlm , for each Enlm .
(c) Enlm does not depend on l and m for the Coulomb potential.
form is given by
Bz Bx iBy
H .
Bx iBy Bz
Eigenvalue of given matrices are given by B and B . H matrices are not diagonals
so e i H t / is equivalent to
iBt
1 e 0
S i Bt
S
0
e
where S is unitary matrices
1 1
2 2
and S 1 S .
1
1
2 2
1 1 i Bt 1 1
iBt 2
e 0
2 e 0
2 2
S 1 i Bt
S , where B / .
0 1 1 i Bt
1 1
e 0 e
2 2 2 2
cos t i sin t
eiHt / , which is equivalent to I cos t i x sin t can be written
i sin t cos t
i B i B
as I cos t sin t , where x
B B
Q4. If the perturbation H ax , where a is a constant, is added to the infinite square well
potential
0 for 0 x
V x
otherwise.
The correction to the ground state energy, to first order in a , is
a a a
(a) (b) a (c) (d)
2 4 2
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Ans: (a)
a2 x a 2 x
Solution: E H ' 0 dx
1 *
0 x sin
2
dx 0 sin .
0
0 0
2
Q5. A particle in one dimension moves under the influence of a potential V x ax 6 , where
a is a real constant. For large n the quantized energy level En depends on n as:
2m 2m
According to W.K.B approximation pdx nh
2m E ax
1/ 2
6
dx n
In the standard basis the matrices for the operators S x1S y2 and S y1S x2 are respectively,
2 1 0 2 1 0 2 i 0 2 i 0
(a) , (b) ,
4 0 1 4 0 1 4 0 i 4 0 i
0 1 0 0 0 i 0 0
0 0 0 i 2 0 0 0 i
2 0 0 i 0 , 0 0 i 0 2 1 0 0 0 2 i 0 0 0
(c) 0 4 0 0 0
(d) ,
4 0 i 0 i 4 0 0 0 i 4 0 0 0 1
i 0 0 0 i
0 0 0
0
0 i 0 0
0 1 0
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Ans: (c)
0 0 0 i
2 0 1 0 i 2 0 0 i 0
Solution: Sx1Sy2
4 1 0 i 0 4 0 i 0 0
i
0
0 0
0 0 0 i
i 0
2
0 1 02
0 i 0
S y1 S x 2 i
4 0 1 0 4 0 i 0 0
i 0 0 0
(B) These two operators satisfy the relation
(a) S x1S y2 , S y1S x2 S z1S z2
(b) S x1S y2 , S y1S x2 0
(c) S S , S S iS S
x
1
y
2
y
1
x
2
z
1
z
2
(d) S S , S S 0
x
1
y
2
y
1
x
2
Ans: (d)
Solution: We have matrix Sx1Sy2 and Sy1Sx2 from question 6(A) so commutation is given by
S S , S S 0 .
x
1
y
2
y
1
x
2
NET/JRF (DEC-2011)
Q7. The energy of the first excited quantum state of a particle in the two-dimensional
potential V x, y
1
2
m 2 x 2 4 y 2 is
3 5
(a) 2 (b) 3 (c) (d)
2 2
Ans: (d)
1 1
Solution: V x, y m 2 x 2 4 y 2 m 2 x 2 m 4 2 y 2 , E n x n y 2
1 1 1
2 2 2 2 2
1 3
For ground state energy n x 0, n y 0 E 2
2 2 2
3 5
First exited state energy n x 1, n y 0
2 2
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Q8. Consider a particle in a one dimensional potential that satisfies V x V x . Let 0
and 1 denote the ground and the first excited states, respectively, and let
(a) 02 0 x 0 12 1 x 1 (b) 0 1 0 x 1 1 x 0
(c) 02 12 (d) 2 0 1
Ans: (b)
Solution: Since V x V x so potential is symmetric.
0 x 0 0 , 1 x 1 0
x 0 0 1 1 0 0 1 1 01 0 x 1 1 x 0
Q9. The perturbation H ' bx 4 , where b is a constant, is added to the one dimensional
m
1/ 4
is 0 e m x
2
/ 2
oscillator potential . You may use the following
1
1
n
integral x e 2n ax 2
dx a 2
n ].
2
Solution: H ' bx 4 , V x
1
m 2 x 2 .
2
m x 2
m
1/ 4
Correction in ground state is given by E 0 H ' 0 1
where 0 e 2
.
0
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1 1
mx 2
m 2
m 2 2 2 mx
2
E bx 0 dx b x e dx b x e
1 * 4 4
dx
0 0
n 1/ 2 1
e dx n 2
2 n x 2
It is given in the equation x
m
Thus n 2 and
1 1 1
2
m 2 2 2 mx m 2 m
1
2 2
E 01 b x e dx b 2
2
1 5
m 2 m 2 5 3 b 2
E b1
.
2 4 m 2 2
0
Q10. Let 0 and 1 denote the normalized eigenstates corresponding to the ground and first
state
1
0 1 , is
2
p i
m
2
a† a 0 , p 2
m † 2
2
a a 2 2 N 1
m †2 m m 1
p2 a a2 2N 1 2N 1 2 1 m
2 2 2 2
p p2 p m
2
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Q11. The wave function of a particle at time t 0 is given by 0
1
u1 u 2 , where
2
u1 and u2 are the normalized eigenstates with eigenvalues E1 and E2
0 is
2 2
(a) (b) (c) (d)
2E 2 E1 E 2 E1 E 2 E1 E 2 E1
Ans: (b)
1
iE1t iE2t
Solution: 0
1
u1 u2 t u1 e
2
u2 e
2
iE1t iE2t
1 1
t is orthogonal to 0 0 t 0 e
e
0
2 2
iE1t iE2t iE1t iE2t E2 E1
i
e e 0e e e
1
cos
E2 E1 t cos t
E2 E1
Q12. A constant perturbation as shown in the figure below acts on a particle of mass m
confined in an infinite potential well between 0 and L .
V0 V0
2
0 L/2 L
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L
V0 2 2 x x
2 L
2
Solution: E 1 V p 1
1
1 sin dx V0 sin 2 dx
0
2 L L L L L
2
V 1 2x 2V 1 2x
2 L
E 0 1 cos
1
dx 0 1 cos dx
L 0 2 L L L 2 L
1
NET/JRF (JUNE-2012)
Q13. The component along an arbitrary direction n̂ , with direction cosines n x , n y , n z , of the
1
spin of a spin particle is measured. The result is
2
(c) n x n y n z
(a) 0 (b) n z (d)
2 2 2
Ans: (d)
0 1 0 i 1 0
Solution: S x , S y , S z
2 1 0 2 i 0 2 0 1
n n x ˆi n y ˆj n z kˆ and n x2 n y2 n z2 1 , S S x iˆ S y ˆj S z kˆ
i
0 0 0
2 +n 2
n S nx 2 n
y
i z
0 0
2 2 2
n
z2
n x in y
nS 2
n x in y
nz
2 2
Let is eigen value of n S
nz
n x in y
2 2
0
n x in y
n z
2 2
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n 2 2 2 2
n n
n x n y2 0 .
2
z z n 2x n 2y 0 z 2
2 2 4 4 4
2 2
4
n x n y2 n z2 2 0 .
2
Q14. A particle of mass m is in a cubic box of size a . The potential inside the box
0 x a,0 y a,0 z a is zero and infinite outside. If the particle is in an
14 2 2
eigenstate of energy E , its wavefunction is
2ma 2
2 14 2 2
2
Solution: E nx ,n y ,nz n x2 n y2 n z2
2ma 2 2ma 2
n x2 n y2 n z2 14 n x 1, n y 2, n z 3 .
Q15. Let nlml denote the eigenfunctions of a Hamiltonian for a spherically symmetric
only of
(a) H , L2 and Lz (b) H and Lz (c) H and L2 (d) L2 and Lz
Ans: (c)
Solution: H En
Q16.
The commutator x 2 , p 2 is
(a) 2ixp (b) 2i ( xp px ) (c) 2ipx (d) 2i ( xp px )
Ans: (b)
Solution: x 2 , p 2 x x, p 2 x, p 2 x xp x, p x x, p p p x, p x x, p px
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9
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x , p xpi xi p pi x i px 2ixp px .
2 2
Q17. A free particle described by a plane wave and moving in the positive z -direction
undergoes scattering by a potential
V , if r R
V r 0
0 , if r R
d 3
2
d 2 2m 2V0 R
2 2
3
2mV0 R 3 d 1
Now V r 2V0 for r R 4 4
d 3 2
3
2
d
Q18. A variational calculation is done with the normalized trial wavefunction
Solution: x
15
5
a 2
x2 , V x 0 , x a and V x , x a
4a 2
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a 2 2
E Hdx where H
a
2m x 2
15 2
a
2 d 2 15 2 15 2
a
a 4a5/ 2
E a x 2
2 5/ 2
a x 2 dx a 2 x 2 2dx
2m dx 4a
5
16a 2m a
a
15 2 2 15 2 4a 3 5 2
E
16a 5 2m a
a 2
x 2
dx
16a 5 m 3
4ma 2
2mV0l 2
2 2
2
1/ 2 o 3
2mV0l 2 2 2
For one bound state V0 . 2
2
8ml 2
2
2
Q20. Which of the following is a self-adjoint operator in the spherical polar coordinate
system r , , ?
i i
(a) (b) i (c) (d) i sin
sin 2 sin
Ans: (c)
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i
Solution: is Hermitian.
sin
NET/JRF (DEC-2012)
Q21. Let v, p and E denote the speed, the magnitude of the momentum, and the energy of a
free particle of rest mass m. Then
Ans: (c)
mv m2v 2 p2v2
Solution: p mv p m v p 2 , m rest mass energy
2 2 2 2
v2 v2 c
1 2 1 2
c c
p2 p2 pc
v 2 m2 2 p 2 v 2 2 2 v
c m c p2 p m2c 2
2
c2
Q22. The wave function of a state of the Hydrogen atom is given by,
200 2 211 3 210 2 211
where nlm is the normalized eigen function of the state with quantum numbers n, l , m in
15 11 3
(a) (b) (c) (d)
6 6 8 8
Ans: (d)
1 2 3 2
Solution: Firstly normalize , 200 211 210 211
16 16 16 16
P0
1 9 10
.
16 16 16
4 2
Probability of getting 1 i.e. P and P .
16 16
Lz
0 1 1
10 4 2 4 2 2
Now, Lz
16 16 16 16 16 16 8
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1 a2 1
(c) En n (d) En n
2 m 2 2
Ans: (a)
p x2 1
Solution: Hamiltonian H of Harmonic oscillator, H m 2 x 2
2m 2
1
Eigenvalue of this, E n n
2
p x2 1 p2 1 2ax a2 a2
But here, H m 2 x 2 ax H x m 2 x 2
2m 2 2m 2 m 2 m 2 4 2m 2
2
p2 1 a a2
H x m 2 x
2m 2 m 2 2m 2
1 a2
Energy eigenvalue, E n n
2 2m 2
Q24. If a particle is represented by the normalized wave function
15 a 2 x 2
, for a x a
x 4a 5 / 2
0
, otherwise
a
15 a 2 x 2 i
p 4a 5/ 2
4a
15 2
5 / 2 xa x 2 dx
a
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a
i a 2 x 2 2 x dx ih 2 155
a
a
15
2
x x3 dx 0 , ( odd function)
16 a
5
a 16a a
2 2
aa x x 2 a x dx
a
15
p 2
2 2 2 2
16a 5
a
2 x3
a
215
2 a 2
x 2
dx 2
15
2 a x
16a 5 a 16a 5 3 a
15 3 2a 3 2 15 2 3 1 15 2 2 5 2
2 2a 5 2a 1
2
16a 5 3 16 a 3 4a 2 3 2a 2
5 2 5
Now, p p2 p 0
2
2a 2 2a
Q25. Given the usual canonical commutation relations, the commutator A, B of
A i xp y yp x and B yp z zp y is
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Solution: If we take S1 S2 S3 1 i.e.
S1 S2 S3
Again S1 S2 S3 1 , then
Energy E 3J
So, minimum energy is 3J and there are two spin configuration.
If we take
S1 S2 S3
2
20 .sinh x 2 0 cosh x 0 sinh x V 0 sinh x 0 sinh x
x x
2 0
2 2 0 coth x 0 V 0 0
x x
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0
2
0 2 0
V 0 2 coth x using relation V 0 4 0
2 0 0 2
x x x
0 d 0
4 0 2 coth x 0 0 2 tanh xdx 0 sec h 2 x .
x 0
NET/JRF (JUNE-2013)
Q28. In a basis in which the z - component S z of the spin is diagonal, an electron is in a spin
1 i / 6
state . The probabilities that a measurement of S z will yield the values
2 / 3
/ 2 and / 2 are, respectively,
(a) 1/ 2 and 1/ 2 (b) 2 / 3 and 1/ 3 (c) 1/ 4 and 3 / 4 (d) 1/ 3 and 2 / 3
Ans: (d)
1 0
Solution: Eigen state of S z is 1 and 2 corresponds to Eigen value and
0 1 2 2
respectively.
2 2
1 1 i
2
2 1 2 2
P , P
2 6 6 3 2 3
b1 0 b2 1
where 0 and 1 denote the ground and first excited states respectively, and b1 and b2
are real constants. The expectation value of the displacement x in the state will be a
minimum when
1 1
(a) b2 0, b1 1 (b) b2 b1 (c) b2 b1 (d) b2 b1
2 2
Ans: (d)
Solution: x b12 0 x 0 b22 1 x 1 2b1b2 0 x 1
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x min b1 b2 b12 b22 0 x 1 b1 b2 1 0 x 1 1 b1 b2 0 x 1 will
2
2 2
so b1 b2
Q30. The un-normalized wavefunction of a particle in a spherically symmetric potential is
given by
r zf r
1
3 211 210 15 211
5
is
(a) 39 2 / 25 (b) 13 2 / 25 (c) 2 2 (d) 26 2 / 25
Ans: (d)
Solution: L2x L2y L2 L2z L2x L2y L2 L2z L2 L2z
9 1 15
L2 L2z = 2 2 1 2 0 2 1 2
25 25 25
24 2 50 24 2 26 2
L2 L2z 2 2
25 25 25
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Q32. Consider a two-dimensional infinite square well
0, 0 x a, 0 ya
V x, y
, otherwise
2 n x x n y y
Its normalized Eigenfunctions are nx ,ny x, y sin sin ,
a a a
where nx , n y 1, 2, 3, ..
a a
V 0 x , 0 y
If a perturbation H ' 0 2 2 is applied, then the correction to the
0 otherwise
V0 V0 64
(a)
4
(b)
4 1 9 2
V0 16 V0 32
(c)
4 1 9 2 (d)
4 1 9 2
Ans: (b)
Solution: For first excited state, which is doubly degenerate
2 x 2 y 2 2 x y
1 sin sin , 2 sin sin
a a a a a a
2 a / 2 2 x 2 a / 2 2 2 y 1 1 V
H11 1 H 1 V0
a 0
sin dx sin
a a 0 a
dy V0 0
2 2 4
2 a/2 x 2 x 2 a / 2 2 y y
H12 1 H 2
a 0
V0
sin
a
sin
a
dx sin
a 0 a
sin
a
dy
4 4 16 16 V
H 12 V0 V0 , H 21 2 H 1 V0 2 and H 22 2 H 2 0 .
3 3 9 9
2
4
V0 16V0
2 2
Thus 4 9 2 0 V0 16V0 0
16V0 V0
4 9
2
9 2 4
V 16V0 V 64
0 0 1 2
4 9 2
4 9
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Q33. The bound on the ground state energy of the Hamiltonian with an attractive delta-
function potential, namely
2 d 2
H a x
2m dx 2
using the variational principle with the trial wavefunction x A exp bx 2 is
Note : e t dt a 1
t a
0
Ans: (a)
A A2
Solution: A, B I and e A 1 .......
1 2
A2 , B A3 , B
A A2
e , B 1
A
......., B = 1, B A, B ....
1 2 2 3
A A, B A, B A A A2 , B A2 , B A
e , B 0 I
A
2!
3!
....
e A , B 1 A
A2
2!
.... e A where A, B I , A 2 , B 2 A and A3 , B 3 A 2 .
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Q35. Two identical bosons of mass m are placed in a one-dimensional potential
V x
1
m 2 x 2 . The bosons interact via a weak potential,
2
V12 V0 exp m x1 x 2 / 4
2
where x1 and x 2 denote coordinates of the particles. Given that the ground state
1
m x 2
m 4
wavefunction of the harmonic oscillator is 0 x e
2
. The ground state
energy of the two-boson system, to the first order in V0 , is
V0
(a) 2V0 (b)
1
2
(c) V0 1 (d) V0 1
2
Ans: (c)
Solution: There are two bosons trapped in harmonic oscillator.
So, energy for ground state without perturbation is, 2 .
2
If perturbation is introduced, we have to calculate V1,2
1
m x12 m x22
m 2
is very tedious task.
But calculating V1,2 on state 0 x 2
e 2
e
So lets use a trick i.e perturbation is nothing but approximation used in Taylor series. So
just expand V1,2 V0 exp m x1 x2 / 4 and take average value of first term
2
m x1 x2 2
V1,2 V0 exp m x1 x2 / 4 V0 1
...
2
4
V0 1
m x12 x22 2 x1.x2
...
4
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m x12 x22 2 x1 . x2 m 0
V1,2 V0 1 ... Vo 1 2m 2m ...
4 4
1 1
2 2
V12 Vo (1 ) V0 1 , so E V0 1 .
4 2 2
NET/JRF (DEC-2013)
1 1 1 i
Q36. A spin - particle is in the state in the eigenbasis of S 2 and S z . If we
2 11 3
h h
measure S z , the probabilities of getting and , respectively are
2 2
1 1 2 9 1 3
(a) and (b) and (c) 0 and 1 (d) and
2 2 11 11 11 11
Ans: (b)
2
1 1 i 1 2
Solution: P 10 2 1
2 11 3 11 11
2
1 1 i 9
P
2
01
11 3 11
i.e. probability of S z getting and
2 2
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1 2 1 2 2 2 2
V x m 2 x 2 x m x 2 x
2 m 2 2 m 2 m 2 4 m 2 4
2
2
V x m 2 x
1
2 m 2 2m 2
1 2 3 1
En n E1 E0
2 2m 2
2 2
Q38. Let nlm denote the eigenfunctions of a Hamiltonian for a spherically symmetric
1
6
200 5 210 10 211 20 211 is
5 5 5
(a) (b) (c) (d)
18 6 18
Ans: (d)
1 5 10 20 10 5
Solution: Lz L z = 0 0 (1) (1) 1
36 36 36 36 36 18
Q39.
If x A exp x 4 is the eigenfunction of a one dimensional Hamiltonian with eigen
value E 0 , the potential V x (in units where 2m 1 ) is
2
x 2
4
4
Ae x VAe x 0 e x 4 x 3 Ve x 0
x
4
4
4
4 3x 2 e x x 3 4 x 3 e x
4
Ve x4
0 12 x 2 e x 16 x 6 e x Ve x 0
4 4 4
V 16 x 6 12 x 2
Q40. A particle is in the ground state of an infinite square well potential is given by,
0 for a x a
V x
otherwise
a a
The probability to find the particle in the interval between and is
2 2
1 1 1 1 1 1
(a) (b) (c) (d)
2 2 2
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Ans: (b)
a a
Solution: The probability to find the particle in the interval between and is
2 2
x x x 1 1 2x
a/2 a/2 a/2
2 2 1
cos cos dx cos 2 dx 1 cos dx
a / 2
2a 2a 2a 2a a / 2
a 2 a a 2 a / 2 2 a
x
a/2
1 1 a a a 1 2a 1 1
1 1
a
x sin a
2a a a / 2 2a 2 2 2a 2
Q41. The expectation value of the x - component of the orbital angular momentum L x in the
state
1
5
3 2,1, 1 5 2,1,0 11 2,1, 1
(where nlm are the eigenfunctions in usual notation), is
(a)
10
25
11 3 (b) 0 (c)
10
25
11 3 (d) 2
Ans: (a)
L L L L
Lx Lx
2 2
1
L 3 2 210 5 2 211
5
1 1 1
L .3 10 110 10(3 11)
25 25 25
1
L 2 5 211 2 11 210
5
1 1
L .3 10 10 11
25 25
L L 1
Lx = 10(3 11)
2 25
Lx
1
25
.3 10
1
25
10 11 =
10
25
11 3
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Q42. A particle is prepared in a simultaneous eigenstate of L2 and Lz . If l l 1 2 and m
are respectively the eigenvalues of L2 and Lz , then the expectation value L2x of the
2 2 2
Ans: (d)
Solution: L2x
1
2
l l 1 2 m 2 2
For max value m 0 and for min m l
l 2 l l 1 2
L2x
2 2
A, B, C are Non zero Hermitian operator.
A, B C AB BA AB Ab 0 C
but C0
if AB BA i.e. A, B C false (2)
NET/JRF (JUNE-2014)
Q43. Consider a system of two non-interacting identical fermions, each of mass m in an
infinite square well potential of width a . (Take the potential inside the well to be zero
and ignore spin). The composite wavefunction for the system with total energy
5 2 2
E is
2ma 2
2 x1 2x 2 2x1 x 2
(a) sin sin sin sin
a a a a a
2 x1 x 2 x x
(d) sin cos sin 2 cos 2
a a a a a
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Ans: (a)
Solution: Fermions have antisymmetric wave function
2 x1 2 x2 2 x1 x2
x1 x2 sin sin sin sin
a a a a a
5 2 2
En nx1 1, nx2 2
2ma 2
Solution: V x, y
1
2
5
m 2 4 x 2 y 2 , E
2
1 1
V x, y m 2 x 2 m 2 y 2
2
2 2
1 1 1 1
Now, E n n x x n y y n x 2 n y
2 2 2 2
3
En 2 n x n y
2
5
En when n x 0 and n y 1 .
2
Q45. A particle of mass m in three dimensions is in the potential
0, ra
V r
, ra
Its ground state energy is
2 2 2 2 3 2 2 9 2 2
(a) (b) (c) (d)
2ma 2 ma 2 2ma 2 2ma 2
Ans: (a)
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d u r l l 1
2 2
Solution: 2
2
V r u r Eu r
2 m dr 2 mr
d 2u r 2mE
K 2u r K , l 0, V r 0
dr 2
2
u r A sin Kr B cos Kr
22
u r A sin Kr , r a, u r 0 sin Ka 0 Ka n E n 1
2ma 2
1
Q46. Given that pˆ r i , the uncertainty p r in the ground state
r r
1
0 r e r / a0 of the hydrogen atom is
a 3
0
2 2
(a) (b) (c) (d)
a0 a0 2a 0 a0
Ans: (a)
1 1
Solution: pˆ r i , 0 r e r / a0 , Pr Pr2 Pr
2
r r a0
3
1 r / a
1 e 0
Now Pr e r / a 0 i 4 r dr
2
0 a 03 r r a 03
4 i r / a0 r / a0 1 1 r / a0 2
r dr
a 03 0
e e
e
a0 r
4 i 1
2 r / a0 2
3
e r dr re 2 r / a0 dr
a 0 a 0 0 0
4 i 1 2! 1!
a03 a0 2 / a0 3 2 / a0 2
4 i a02 a02
0
a03 4 4
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1
2
2 r / a0
Pr2 3 e r / a0 2 2 e 4 r dr
2
a 0 0 r r r
4 2 r / a0 r / a0 1 2 1 r / a0 2
a03 0
e e a 2 r a e r dr
0 0
4 2 1 2 2 r / a0
2
4 2 1 2 ! 2 1 !
a03 0 a02 a0 0
2 r / a0
r e dr re dr 2
a0 2 / a0 a0 2 / a0
3 2
a03
4 2 2 ! a03 2 a02 4 2 a0 a0 4 2 a0 2
4
a03 a02 8 a0 4 a03 2 a03 4 a02
2
P Pr2 Pr 0
2
2
a0 a0
Q47. The ground state eigenfunction for the potential V x x where x is the delta
function, is given by x Ae
x
, where A and 0 are constants. If a perturbation
H bx 2 is applied, the first order correction to the energy of the ground state will be
b b 2b b
(a) (b) (c) (d)
2 2
2
2
2 2
Ans: (d)
Solution: V x x , x Ae
x
1 x e x
E 1 H 1 e x bx 2 e x dx
1
1
0
e bx dx b e x dx b x e dx x e dx b 2 x e dx
2 x 2 2 x 2 2 2 x 2 2 x 2 2x
0 0
2! 2! b
2 x
e bx 2
dx 2b 3
2 b 3
2 8 2 2
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Q48. An electron is in the ground state of a hydrogen atom. The probability that it is within the
Bohr radius is approximately equal to
(a) 0.60 (b) 0.90 (c) 0.16 (d) 0.32
Ans: (d)
2
4
a0 a0
1
4 r dr 3 r e
r / a0 2 2 2 r / a0
Solution: Probability: e dr
0 a 3
0
a0 0
4 2 2 r / a0 a0 2 r / a0 a0 a0 a0
a0 a0 a0
2 r / a0 a0 a0
a03
r e
2
0
2 r e
2 2
0
2 e
2 2 2
0
Ans: (c)
3x
0, 0 xa A sin , 0 xa
Solution: V x x a
, otherwise 0 , otherwise
x x 3 x
x A sin 3
a
3
A sin
4 a
1
A sin
4 a
sin 3 A 3sin A 4sin A 3
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A a 2 x a 2 3 x A a a
3sin sin x 3 1 x 3 x
4 2 a a 2 a a 4 2 2
a 2 a 2 10a 2 32
1 9 A A 1 A 1 A
32 32 32 10a
1 a 32 a 32 3 1
x 3. 1 x 3 x 1 x 3 x
4 2 10a 2 10a 10 10
22 9 2 2
Now, E1 2
, E3 E an P an
2ma 2ma 2
2 2
1 9 2 1
Probability P E1 , P E3
10 10
9 2 2 1 9 2 2 9 2 2
E E
10 2ma 2 10 2ma 2 10ma 2
NET/JRF (DEC-2014)
Q50. Suppose Hamiltonian of a conservative system in classical mechanics is H xp , where
is a constant and x and p are the position and momentum respectively. The
corresponding Hamiltonian in quantum mechanics, in the coordinate representation, is
1 1
(a) i x (b) i x
x 2 x 2
i
(c) ix (d) x
x 2 x
Ans: (b)
Solution: Classically H xp , quantum mechanically H must be Hermitian,
So, H xp px and H xp px
2 2
i x
H x i i x x
2 x x 2 x x
i 1
H 2x i x
2 x x 2
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Q51. Let 1 and 2 denote the normalized eigenstates of a particle with energy eigenvalues
E1 and E2 respectively, with E 2 E1 . At time t 0 the particle is prepared in a state
t 0
1 2
1
2
The shortest time T at which t T will be orthogonal to t 0 is
2
(a) (b) (c) (d)
E 2 E1 E 2 E1 2E 2 E1 4E 2 E1
Ans: (b)
1 iE1T 1 iE2T
1 2
1
Solution: t 0 and t T e 1 e 2
2 2 2
1 iE1 T 1 iE2 T iE
1T
iE
2T i E2 E1
T
0 T dx 0 e e 0 e e e 1
*
2 2
T
Equate real part cos E2 E1 1 T cos 1 1
E2 E1 E2 E1
Q52. Consider the normalized wavefunction
a1 11 a 2 10 a3 11
where lm is a simultaneous normalized eigenfunction of the angular momentum
1 1 1 1
(a) a1 a3 , a2 (b) a1 a3 , a2
2 2 2 2
1 1 1
(c) a1 a3 , a 2 (d) a1 a 2 a3
2 2 3
Ans: (b)
L L
Solution: Lx
2
For L , L a1 11 a2 10 a3 11 a1 0 12 a2 2 11 a3 2 10
a2 2 11 a3 2 10
L L
Given
2
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L L 1
a2 2 11 a1 a3 2 10 a2 2 11
2 2
L L
a1 11 a2 10 a3 11 (Given)
2
a2
Thus a1 a2 2a1
2
a1 a3 a1 a3 a22
a2 2a1 a1 a3 a12 a22 1
2 2 2
1 1
a1 a3 , a2
2 2
Q53. Let x and p denote, respectively, the coordinate and momentum operators satisfying the
canonical commutation relation x, p i in natural units 1 . Then the commutator
x, pe is
p
Ans: (b)
Solution: a a1iˆ a2 ˆj a3 kˆ , b b1iˆ b2 ˆj b3 kˆ , x iˆ y ˆj z kˆ
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a , b a1 x a2 y a3 z , b1 x b2 y b3 z
a , b a1b1 x , x a1b2 x , y a1b3 x , z a2b1 y , x a2b2 y , y
a1b1 0 a1b2 2i z 2ia1b3 y a2b1 2i z 0 a2b3 2i x a3b1 2i y a3b2 2i x 0
a , b 2i a b
Q55. The ground state energy of the attractive delta function potential
V x b x ,
where b 0 , is calculated with the variational trial function
x
A cos , for a x a,
x 2a is
0, otherwise,
mb 2 2mb 2 mb 2 mb 2
(a) (b) (c) (d)
2 2 2 2 2 2 2 4 2 2
Ans: (b)
x
Solution: V x b x ; b 0 and x A cos ; a x a
2a
2 x
Normalized cos
2a 2a
a 2 2 22
T * 2 dx
a
2m x 8ma 2
2 b
V * b x dx b
a
a 2a a
2 2 b E 2 2 2 b 2 2 2 2
E 0 b 0 a
8ma 2 a a 8ma 3 a2 4ma 4mb
b 4mb b 4mb
2
22
2 2
2mb 2
Put the value of a in equation: E
8ma 2 a 8m 2 2
2
2 2
2 2
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Q56. Let c0 0 c1 1 (where c0 and c1 are constants with c02 c12 1 ) be a linear
combination of the wavefunctions of the ground and first excited states of the one-
dimensional harmonic oscillator. For what value of c0 is the expectation value x a
maximum?
1 1
(a) x , c0 (b) x , c0
m 2 2m 2
1 1
(c) x , c0 (d) x , c0
2m 2 m 2
Ans: (c)
Solution: c0 0 c1 1
X X
X 2c0 c1 0 X 1 c02 c12 c0 c1 0 X 1 1 c0 c1 0 X 1
2
2
1
For max X c0 c1 c02 c12 1 c0
2
1 1
X 2 0 X 1 0 X 1
2 2
2m
0 aa 1
X
2m
Q57. Consider a particle of mass m in the potential V x a x , a 0 . The energy eigen-
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2 1
According to W.K.B., pdq n where a1 and a2 are positive mid point
1 2
P2
E a x P 2m E a x
2m E / E /
1
2m E a x dx n
E/ a
E/ a
2
1
2m E ax dx 2m E ax dx n
0 E/a
E / a 0
2
1
2m E ax dx n
E/a
2
0
2
2m E ax t , At x 0, t 2mE; x E / a, t 0
2madx dt
2 mE
2 mE 1 2 1
2ma t dt n 2ma t 3/ 2
1/ 2
n
0
2 3 0 2
4 1 4 1
n ma 2mE n
2 mE
3/ 2
ma t 3/2
3 0
2 3 2
2/3
4 1 3a 1
23/ 2 am5 / 2 E 3/ 2 n E n
3 2 4 2m 2
Q58. The Hamiltonian H 0 for a three-state quantum system is given by the matrix
1 0 0 0 1 0
H 0 0 2 0 . When perturbed by H 1 0 1 where 1 , the resulting shift
0 0 2 0 1 0
in the energy eigenvalue E 0 2 is
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2 0 0 1
0 2 in H 0 is not 0 1 0 in H because H is not in block diagonal form. So we
0 0 0
After diagonalisation H 0 0 2 0 , 0 is correction for Eigenvalue at H 0 .
0 0 2
NET/JRF (JUNE-2015)
Q59. The ratio of the energy of the first excited state E1 , to that of the ground state E0 , to that
L
of a particle in a three-dimensional rectangular box of side L, L and , is
2
(a) 3 : 2 (b) 2 :1 (c) 4 :1 (d) 4 : 3
Ans. (a)
22 6 2 2
Solution: E n 2
n 2
4 n 2
, for ground state n 1, n 1, n 1 E
2mL2
x y z x y z 0
2mL2
22 9 2 2
For first excited state nx 1, n y 2, nz 1 E E1 1 4 4
2mL2 2mL2
E1 9 3
E0 6 2
Q60. If Li are the components of the angular momentum operator L , then the operator
i 1,2,3 L, Li equals
(a) L (b) 2L (c) 3L (d) L
Ans. (b)
Solution: Let L Lx iˆ Ly ˆj Lz kˆ
x 1, y 2, z 3
L, Lx Ly , Lx j Lz , Lx kˆ iLz ˆj Ly kˆ i
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L, Lx , Lx i Lz , Lx ˆj Ly , Lx i i .i Ly ˆj i Lz i Lz i .kˆ 2 Ly ˆj Lz kˆ
similarly, L, Ly Ly 2 Lx iˆ Lz kˆ
L, Lz Lz 2 Lx iˆ Ly ˆj
L, Li Li 2 2 Lx iˆ Ly ˆj Lz kˆ 2 L put 1
i 1,2,3
T x x a
ipx ipx ipa ip x a
T p T x e
dx x a e
dx e
x a e
dx
ipa
T p e p
1 1 1
(c) 4 N a 2 b 2 (d) 4 N a 2 b 2
2 3 3
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Ans. (d)
d
Solution: a 2 b 2 cos 2
d
2 2
2 b2
a 2 sin d d b 2 cos 2 sin d d a 2 .4 b 2 .2 4 a 2
0 0 0 0
3 3
b2
Number of particle scattered per unit time, .N 4 N a 2
3
1
Q63. A particle of mass m is in a potential V m 2 x 2 , where is a constant. Let
2
m ipˆ daˆ
aˆ xˆ . In the Heisenberg picture is given by
2 m dt
daˆ 1 m p 2 im 2 1 m 2 p i
x, pˆ , x 2 i 2 x i
dt i 2 2m 2m i 2 2m 2
m p m ip
i x i x i aˆ
2 m 2 m
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Q64. Two different sets of orthogonal basis vectors
1 0 1 1 1 1
, and , are given for a two dimensional real vector space.
0 1 2 1 2 1
The matrix representation of a linear operator  in these basis are related by a unitary
transformation. The unitary matrix may be chosen to be
0 1 0 1
(a) (b)
1 0 1 0
1 1 1 1 1 0
(c) (d)
2 1 1 2 1 1
Ans. (c)
1 0 1 1 1 1 1 1
Solution: u1 , u2 u u1 u2
0 1 2 1 1 2 1 1
Q65. The Dirac Hamiltonian H c . p mc 2 for a free electron corresponds to the classical
relation E 2 p 2 c 2 m 2 c 4 . The classical energy-momentum relation of a piratical of
q
2
2
charge q in a electromagnetic potential , A is E q c p A m 2 c 4 .
2
c
Therefore, the Dirac Hamiltonian for an electron in an electromagnetic field is
e e
(a) c . p A. A mc 2 e (b) c . p A mc 2 e
c c
e e
(c) c . p e A mc 2 (d) c . p A mc 2 e
c c
Ans. (d)
Solution: Electromagnetic interaction of Dirac particle
1
qA 2 2
H P c m c q
2 2 4
c
Quantum mechanical Hamiltonian
qA
i c P mc q
2
t c
put q e
e
H c . P A mc 2 e
c
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Q66. A particle of energy E scatters off a repulsive spherical potential
V for r a
V r 0
0 for r a
where V0 and a are positive constants. In the low energy limit, the total scattering cross-
2
1 2m
section is 4 a tanh ka 1 , where k 2 2 V0 E 0 . In the limit V0
2
ka h
the ratio of to the classical scattering cross-section off a sphere of radius a is
1
(a) 4 (b) 3 (c) 1 (d)
2
Ans. (a)
2
1
Solution: 4 a tanh ka 1
2
ka
2
1
ka , tanh ka 1 4 a 1 2
ka
and ka , lim H 4 a 2
ka
H
classically c a 2 4
c
NET/JRF (DEC-2015)
Q67. A Hermitian operator O has two normalized eigenstates 1 and 2 with eigenvalues 1
and 2 , respectively. The two states u cos 1 sin 2 and v cos 1 sin 2
are such that v O 7 / 4 and u v 0 . Which of the following are possible values of
and ?
(a) and (b) and
6 3 6 3
(c) and (d) and
4 4 3 6
Ans.: (a)
Solution: u cos 1 sin 2 , v cos 1 sin 2
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7
it is given Oˆ 1 1 , Oˆ 2 2 2 v Oˆ v
4
7 7
cos 2 2sin 2 cos 2 sin 2 1 sin 2 1
4 4
3
sin
2 3
u v 0 cos cos sin sin 0 cos 0
5
or or or
2 2 2 3 3 2 6 6
x
Q68. The ground state energy of a particle of mass m in the potential V x V0 cosh ,
L
2
where L and V0 are constants (with V0 ) is approximately
2mL2
2V0 V0
(a) V0 (b) V0
L m L m
V0 V0
(c) V0 (d) V0
4L m 2L m
Ans.: (d)
x V
Solution:
V0 cosh 0 e x / L e x / L
L 2
V0 x 1 x V0 x 1 x
2 2
1 .... 1 ....
2 L 2! L 2 L 2! L
2
V V V x 1 V
0 0 0 V0 02 x 2
2 2 2 L 2 L
V0 V0
K ,
L2 mL2
So, ground state energy is
V0 V0
V0 V0 2
V0
2 2 mL 2L m
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Q69. Let nlm denote the eigenstates of a hydrogen atom in the usual notation. The state
1
2 200 3 211 7 210 5 211
5
is an eigenstate of
(a) L2 , but not of the Hamiltonian or Lz (b) the Hamiltonian, but not of L2 or Lz
(c) the Hamiltonian, L2 and Lz (d) L2 and Lz , but not of the Hamiltonian
Ans.: (b)
1
Solution: 2 200 3 211 7 2 10 5 2 1 1
5
13.6
H
4
So is eigen state of H
But L2 and Lz
1
Q70. The Hamiltonian for a spin- particle at rest is given by H E0 z x , where x
2
and z are Pauli spin matrices and E0 and are constants. The eigenvalues of this
Hamiltonian are
(a) E0 1 2 (b) E0 1 2
1
(c) E0 (doubly degenerate) (d) E0 1 2
2
Ans.: (a)
1 0 0 1 1
Solution: H E0 z x E0 H E0
0 1 1 0 1
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Q71. A hydrogen atom is subjected to the perturbation
2r
V pert r cos
a0
where a0 is the Bohr radius. The change in the ground state energy to first order in
(a) (b) (c) (d)
4 2 2 4
Ans.: (d)
Solution: For First order perturbation
r
1 2r
E11 100 V p 100 , 100 e a , V p cos
a03 a0
2 r 2 r
1 2r 4 2r
E 3 e a0 cos 4 r 2 dr 3 e a0 cos r 2 dr
1
1
0
a0 a0 a0 0 a0
ia2 r i 2 r
2 a0 1i 2
2 r 2 r 2 r 1 i
4 e e 0
0 a
2
3 e
a03 0 0
a0
r dr e r dr e a0
r 2
dr
a0 0
2
2 2! 2! 1 1
a03 2
3
3
2 1 i 3 1 i 3
1 i 1 i
2
a0 0
a
1 1 1 1
2 3
3 1 i
3
4 2 4
i 3 i 3
1 i
3
2 2 e e 4
2 2
i 34 i 3
3
e e 4
2 cos 4
4 2 4 2
1
2 E11
4 2 2 4 4
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Q72. The product of the uncertainties Lx Ly for a particle in the state a 1,1 b 1, 1
3 1
(c) a and b (d) a b
2 2
Ans.: (d)
Solution: a 1,1 b 1, 1 , L 2b 1, 0 , L2 2 2b 1,1
L 2a 1, 0 , L2 2 2 a 1, 1
L2 a 2 2 b 2 2 a b 2 2 2
2 2
2
L2z a b 2 2 2
Lx 0, Ly 0
L2x
1 2
4
L L2
2 L2
L2
2 4
1 a*b b*a 2 2 2 2 2 2 a 2
b
2
2 *
L2x
2
a b b* a a b
2 2
L2y
2 L2 L22 L2 L2
4
2
2
L2y
2
a b a*b b*a
2
2 2
a b b a
2 2
Lx Ly a b
2 2 2
* *
a b 1
2
2
2
Lx Ly 1 a*b b*a (i)
2
1 i 2
Now check option (a) a ib a ,b Lx Ly
2 2 2
2
Option (b) a 0, b 1 Lx Ly
2
3 1 2
Option (c) a ,b Lx Ly
2 2 4
1 1
Option (d) a b a ,b Lx Ly 0 option (d) is correct
2 2
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Q73. The ground state energy of a particle in potential V x g x , estimated using the trail
wavefunction
c 2
x a5
a x2 , x a
0, x a
(where g and c are constants) is
1/ 3 1/ 3 1/ 3 1/ 3
15 2 g 2 5 2 g 2 3 2 g 2 7 2 g 2
(a) (b) (c) (d)
16 m 6 m 4 m 8 m
Ans.: (a)
a
15
dx 1 c 16
*
Solution:
a
2 15 2 2
a
10 2
T
2m 16a 2 a
a 2
x 2
x 2
a
x 2
dx T
4ma 2
15 2 g
a
V
16a 05 5
x a 2 x 2 dx V ga
16
E T V (i)
10 2 5 ga
E
4ma 2 16
1
dE 8 2 3
0 a3 a 2
da mg mg
put the value of a in equation (i)
1
15 2 g 2 3
E
16 m
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NET/JRF (JUNE-2016)
Q74. The state of a particle of mass m in a one dimensional rigid box in the interval 0 to L is
2 3 2 x 4 4 x
given by the normalized wavefunction x sin sin . If its
L 5 L 5 L
energy is measured the possible outcomes and the average value of energy are,
respectively
h 2 2h 2 73 h 2 h2 h2 19 h 2
(a) , and (b) , and
2mL2 mL2 50 mL2 8mL2 2mL2 40 mL2
h 2 2h 2 19 h 2 h 2 2h 2 73 h 2
(c) , and (d) , and
2mL2 mL2 10 mL2 8mL2 mL2 200 mL2
Ans: (a)
2 3 2 x 4 4 x
Solution: x sin sin
L 5 L 5 L
n 2 2 2
Measurement E
2mL2
h2 2h 2
n 2 E2 and n 4 E4
2mL2 mL2
9 16
Probability p E2 and p E4
25 25
Now, average value of energy is
9 h2 16 2h 2 73h 2
E an p an
25 2mL2 25 mL2 50mL2
Q75. If Lˆ x , Lˆ y , Lˆ z are the components of the angular momentum operator in three dimensions
(a) iLx Lˆ2z Lˆ2y (b) iLˆ z Lˆ y Lˆ x
(c) iL 2 Lˆ Lˆ
x
2
z
2
y (d) 0
Ans: (a)
Solution: Lx , Lx Ly Lz Lx Lx , Ly Lz Lx , Lx Ly Lz
Lx Lx , Ly Lz Lx Ly Lx , Lz 0 Lx iLz Lz Lx Ly iLy
iLx L2z iLx L2y iLx L2z L2y
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Q76. Suppose that the Coulomb potential of the hydrogen atom is changed by adding an
Ze 2 g
inverse-square term such that the total potential is V r 2 , where g is a
r r
constant. The energy eigenvalues Enlm in the modified potential
(a) 1 (b)
E sin 1
2
E2 cos 2
E12 E22
(c) eiE1t / sin eiE2t / cos (d) e iE1t / sin 2 eiE2t / cos 2
Ans: (a)
Solution: t 0 sin 1 cos 2
iE1t iE2t
t sin 1 e
cos 2 e
i E1 E2 t
t t sin 1 1 cos 2 2 2 Re e
2 2
sin cos 1 2
sin 2 cos 2 0 1 1 2 0
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1
Q78. Consider a particle of mass m in a potential V x m 2 x 2 g cos kx . The change in
2
1
the ground state energy, compared to the simple harmonic potential m 2 x 2 , to first
2
order in g is
k 2 k 2
(a) g exp (b) g exp
2m 2m
2k 2 k 2
(c) g exp (d) g exp
m 4m
Ans: (d)
Solution: Ground state wavefunction
1
m x
m 4 2
2
0 x e
The perturbation term is H p g cos kx
First order correction E01 0* x H P 0 x dx
1 1
g m 2 m x ikx
m x 2 m x
m 2 eikx e ikx
2 2
ge dx e .e dx e . e ikx dx
2 2
1 1
m x m x
g m 2 g m 2
2 2
ikx ikx
e dx e
dx
2 2
From 1st term, we have
m 2 2 ikx ik ik
1 2 2 1
m
2
ik
g m 2 x
2 m 2 m 2 m g m 2 x
k 2
e
2 m
e dx e 4 m
dx
2 2
1
m
2
ik
g k m 2
2
k 2
x
e
2 m
e 4 m dx e 4 m
2
1
m x
g m 2
2
ikx
e
Similarly, from term (ii), dx
2
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47
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1
m
2
ik
g 4kmh m 2
2
k 2
x
2 m
e e dx e 4 m
2
g k k
2 2 2
k
Hence, E e 4 m e 4 m ge 4 m
1
0
2
Q79. The energy levels for a particle of mass m in the potential V x x , determined in
1
b
2m E V x dx n
a 2
E E
1 1
0
2m E xdx E xdx n 2 2m E x dx n
E 0 2 0 2
dt
put E x t , dx
E
limit x 0 t E , x t 0
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48
fiziks
Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
dt 1
0
2 2m t n
E 2
0
2 2m 2 32 1 2 2m 2 32 1
t n .E n h
3 E 2 3 2
2
1 3
3
3 1 3
E n 2
En n
2 4 2m 4 2m 2
Q80. A particle of mass m moves in one dimension under the influence of the potential
V x x , where is a positive constant. The uncertainty in the product
(a) 2 (b) (c) (d) 2
2 2
Ans: (c)
Solution: V x x
0
2 2
2
0
d2 x d
2
x d
2
p 2 2 * 2
dx 2
e 2
ex
dx 2
e 2
e x dx
dx
dx 0
dx
2 3 2 3
0
e e
2 x 2 x
2 3
dx 2 3
dx 2 2 , which is not possible
0
2 2
2
d
dx 2 2 , p p2 p
2 2 2
so, we will use the formula p
dx
1
now, x.p .
2 2
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49
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2 4
Q81. The ground state energy of a particle of mass m in the potential V x x ,
6m
1
x 2
4
estimated using the normalized trial wavefunction x e 2
, is
1 3
dxx 2 e x dx x 4 e x
2 2
[use and ]
2 4 2
3 2 13 8 2 13 2 2 13 3 2 13
(a) (b) (c) (d)
2m 3m 3m 8m
Ans: (d)
1
x
4 2
2
Solution: E T V , for x e 2 , T
4m
1 1
2 4 x 2 2 3 2
2 2
V x e dx x 4 e x dx . 2
2 2
6m 6m
6m 4 8m 2
2 2
E (i)
4m 8m 2
dE 2 2 2 2 1
0 1 0 3
d 4m 8m 3 4m 3
Putting the value of in equation (i),
1
2 1
2 2 1
3 3 2 13
E 3
3
2
4m 2 8m
8m 3
4m
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
NET/JRF (DEC-2016)
Q82. Consider the two lowest normalized energy eigenfunctions 0 x and 1 x of a one
d 0
dimensional system. They satisfy 0 x 0* x and 1 x , where is a real
dx
constant. The expectation value of the momentum operator in the state 1 is
2
(a) (b) 0 (c) (d)
2
2
2
Ans. : (b)
d 0
Solution: 1 x
dx
d 0 d 2
px px dx 1 i 1 dx * i 20 dx
*
*
x
1
dx dx
d 0 d 2 0
i
2
dx
dx dx 2
Integrate by parts
2 d d 0 d 2 0 d 0
d 0 d 2 0
I i 0 dx 2 dx dx 0 i dx dx 2 dx
2
dx dx
d 0 d 2 0
I 0 i dx dx 2 dx
2
d 0
0 , 0 0, x
dx
I 0 I 2I 0 I 0 px 0
d
Q83. Consider the operator, a x acting on smooth function of x . Then commutator
dx
, cos x is
(a) sin x (b) cos x (c) cos x (d) 0
Ans. : (a)
d
Solution: a x
dx
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51
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d cos xd
cos x sin x sin x
dx dx
a, cos x x sin x
a, cos x sin x
Q84. Consider the operator p qA , where p is the momentum operator,
A Ax , Ay , Az is the vector potential and q denotes the electric charge. If
B Bx , By , Bz denotes the magnetic field, the z -component of the vector operator
is
(a) iqBz q Ax p y Ay px (b) iqBz q Ax p y Ay px
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52
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Q85. The 2 2 identity matrix I and the Pauli matrices x , y , z do not form a group under matrix multiplication. The minimum number of 2 2
matrices, which includes these four matrices, and form a group (under matrix multiplication) is
(a) 20 (b) 8 (c) 12 (d) 16
Ans. (d)
I I i I i I x x i x i x y y i y i y z z i z i z
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53
fiziks
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H.No. 40‐D, Ground Floor, Jia Sarai, Near IIT, Hauz Khas, New Delhi‐110016
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54
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
Q86. The dynamics of a free relativistic particle of mass m is governed by the Dirac
Hamiltonian H c . p mc 2 , where p is the momentum operator and x , y , z
and are four 4 4 Dirac matrices. The acceleration operator can be expressed as
2ic
(a) cp H (b) 2ic 2
ic 2ic
(c) H (d) cp H
Ans. : (a)
Solution: H c . p mc 2
If vx velocity of x direction
From the Ehrenfest theorem
dx 1 x 1
vx x, H x, c x px c y p y c z p z mc 2 0
dt i t i
c
x, x px c x
i
Similarly, acceleration is given by
dvx 1 c
ax c x , H x , c x px c y p y c z pz mc 2
dt i i
Using relation i j j i 0 , i i 0 and i , p j 0
x , c x px 0
x , c y p y c x y y x p y y c x , p y c x y c x y p y 0 2c x y p y
x , c z pz c x z c z x pz z c x , pz c x z c x z pz 0 2c x z pz
x , mc 2 x x mc 2 2mc 2 x
c
ax 2c x y p y 2c x z p z 2 x mc 2
i
2 x c
ax c y p y c z pz mc 2 c x p x c x px
i
2 x c
ax c x p x c y p y c z pz mc 2 c x px
i
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2c
ax x .H c x x px
i
2ic
c x x px x .H , x2 x
2ic
a ax iˆ a y ˆj az kˆ
cp .H
Q87. A particle of charge q in one dimension is in a simple harmonic potential with angular
2
t
frequency . It is subjected to a time- dependent electric field E t Ae
, where A
and are positive constants and 1 . If in the distant past t the particle was
in its ground state, the probability that it will be in the first excited state as t is
proportional to
1 1
2 2 1
2
(a) e (b) e 2 (c) 0 (d)
2
Ans. : (a)
2
t2
i fi t
Solution: Transition probability is proportional to Pif e 2
e where
3 1
fi 2 2
2
t2
Pif exp it dt
2
1 2 2 2
i 2
2
t2
2 i
Now calculate exp 2 it dt exp 2 t it
2 2
2 2 1 it
2
exp exp 2
t dt
4 2
2
t2
Pif exp it dt
2
2
2 2 1 it
2
Pif exp exp 2 t dt
4 2
2 2
Pif exp
2
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Q88. A particle is scattered by a central potential V r V0 re r , where V0 and are positive
constants. If the momentum transfer q is such that q q , the scattering cross-
dn
x e dx n
n ax
[You may use e ax dx ]
da
(a) q 8 (b) q 2 (c) q 2 (d) q 6
Ans. : (a)
Solution: The form factor is given for high energy as q
2 m 2m
f , rV r sin qr dr 2 r 2V0 e v sin qr dr
q 0
2
q 0
mV0 2 r iq
2 m 2 r e
iqr
e iqr
2 V0 r e dr 2 i r e dr r 2 e dr
r iq
q 0 2i q 0 0
mV0i
2
2
2
2 0
2mV i iq iq
3 3
q iq 3 iq 3 q iq 3 iq 3
2mV0 i iq 3 iq 3 q iq 3 iq 3 q
3 3 2 2 3 3 2 2
2
q 2 q2
3
3 2
2mV0i 6 2iq 2iq 3 2mV0 2q 6 q
2 2
q 2 q 2 3
q 2 q2 3
q3 6 2 1 1 1 2
2 2 3
q2 6
4 2 1
q q 2 q q q
q 2 1
6
q
f q 4 q 8
2 2
2 d x
2
Solution: A x a x E x
2m dx 2
Integrates both side within limit
a to a
a a a
2 d 2
dx A x a dx E x dx
2m a dx 2
a a
2 d II d I
A a 0
2m dx dx
d II d I 2mA
2 a
dx dx
d 2mA
so discontinues in at x a is 2 a .
dx
NET/JRF (JUNE-2017)
Q90. If the root-mean-squared momentum of a particle in the ground state of a one-
dimensional simple harmonic potential is p0 , then its root-mean-squared momentum in
the first excited state is
(a) p0 2 (b) p0 3 (c) p0 2 / 3 (d) p0 3 / 2
Ans. : (b)
Solution: P m Pˆ m
a a†
2i
m 2
P2
2
a a†2 2 N 1
m
P2
2
a 2 a †2 2 N 1
For any state n ,
a 2 0, a †2 0 and 2 N 1 2n 1
m
P 2 2n 1 and P 0
2
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m
Prms P 2 P Prms 2n 1
2
m
For ground stat n 0, Prms P0
2
m
So, for n 1, Prms 3
2
Prms 3P0
Q91. Consider a potential barrier A of height V0 and width b , and another potential barrier B
of height 2V0 and the same width b . The ratio TA / TB of tunnelling probabilities TA and
(a) exp 1.99 0.99 8 mV0b 2 / 2
(b) exp 1.98 0.98
8 mV0b 2 / 2
(c) exp
2.99 0.99 8 mV0b 2 / 2
(d) exp
2.98 0.98 8 mV0b 2 / 2
Ans. : (a)
2 m V E V0
Solution: T e , where E
100
For potential A, V V0
2m V0
V0
TA e 2 100
2 m 99
V0
2 m 0.99V0
TA e 2 100
e
V0
For Potential B, V 2V0 and E =
100
2m V0
2V0
TB e 2 100
2 m 199V0
2 m 1.99V0
TB e 2 100
e
0.99V0
TA e
TB e 1.99V0
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TA
TB
e 1.99V0
e
0.99V 0
Q92. A constant perturbation H is applied to a system for time t (where H t )
leading to a transition from a state with energy Ei to another with energy E f . If the time
at ti 2ti ,
2
4 f v i fi ti
pif sin 2
h 2 2
fi 2
at ti 2ti ,
2 2
4 f v i fi 2ti 4 f v i
p ff sin 2 sin fi ti
h 2 2fi 2 h 2 2fi
sin 2 fi ti
2fi ti2
pif sin fi ti
2
t 2 2
t1 0
fi i
p ff t t t
2
sin fi i
2
sin 2 fi i fi i
2 2 2
2fi ti2
2
4 2fi ti2
4
2fi ti2
pif (2)
4 pif (2) 4 pif (1)
pif (1)
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Institute for NET/JRF, GATE, IIT‐JAM, M.Sc. Entrance, JEST, TIFR and GRE in Physics
a b
Q93. The two vectors and are orthonormal if
0 c
(a) a 1, b 1/ 2, c 1/ 2 (b) a 1, b 1, c 0
(c) a 1, b 0, c 1 (d) a 1, b 1/ 2, c 1/ 2
Ans. : (c)
a b
Solution: 1 , 2
0 c
1 1 1 a 1
2 2 1 b c 1
2 2
1 2 0 a 0 bc 0
a.b 0 c 0 a b 0
so b 0
c 1, c 1
2
a 1, b 0, c 1
Q94. Consider the potential
V r i V0 a 3 r ri
3
where ri are the position vectors of the vertices of a cube of length a centered at the
2
origin and V0 is a constant. If V0 a 2 , the total scattering cross-section, in the low-
m
energy limit, is
2
mV0 a 2
2 16a 2 mV0 a 2
(a) 16a (b)
2
2 2
2
64a 2 mV0 a 2 64a 2 mV0 a 2
(c) (d)
2 2 2
Ans. : (c)
Solution: V r V0 a 3 3 r ri
i
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V0 a x xi y yi z zi
3
m
f 2
V r d 3 r
2
m 8
x xi y yi z zi dxdydz
3
V a
2 2
0
i 1
m
V a 3 1 1 1 1 1 1 1 1
2 2 0
8mV0 a 3 4mV0 a 3
2 2 2
16m 2V02 a 6
Differential scattering cross section D f
2
24
16m 2V02 a 6 64a 2 m 2V02 a 4
4
24 h4
2
64a 2 mV0 a 2
2
Ans. : (b)
Solution: H ' bx 2 put x r sin cos
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1 H 1r sin dr d d
* 2
2r 2
b
3 r 2 e a0 r 2 dr sin 3 d cos 2 d ba02
a0 0 0 0
5 1 5
(a) (b) (c) (d)
14 2 7
Ans. : (b)
A a 2 x 2 , a x a
Solution: x
0 , otherwise
For normalization
dx 1
*
15 15
A2 5
A
16a 16a 5
2 2 2 15
a a
T * dx 2 2 a 2 x 2 dx
5
2m a x 2
2m 16a 0
5 2
T
4ma 2
a a
2 x 2 a 2 x 2 dx.
1 1 15
V *V dx , where V x 5
2
m 2 x 2 m 2
a
2 2 16a 0
m 2 a 2
V
14
5 2 m 2 a 2
E T V
4ma 2 14
dE 5 2 2 m 2 a 35 2
0 0 a 4
.
da 4ma 3 7 2 m 2 2
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1/ 2
35
a
2
.
2 m
5 2 m 2 m 2 35
E . .
4 m 35 14 2 m
5 2 1 35 5 5 5
2 2 35 7 2 2 14 14 14
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THERMODYNAMICS AND STATISTICAL PHYSICS
NET/JRF (JUNE-2011)
Q1. Consider the transition of liquid water to steam as water boils at a temperature of 1000 C
under a pressure of 1 atmosphere. Which one of the following quantities does not change
discontinuously at the transition?
(a) The Gibbs free energy (b) The internal energy
(c) The entropy (d) The specific volume
Ans: (a)
Solution: In first order transition Gibbs free energy is continuous.
Q2. A particle is confined to the region x 0 by a potential which increases linearly as
u x u 0 x . The mean position of the particle at temperature T is
k BT k BT
(b) k B T / u 0
2
(a) (c) (d) u 0 k B T
u0 u0
Ans: (a)
p2 u x
1 0
Solution: Partition function Z e 2 mkBT dp e kBT dx and x xp x dxdpx
h
0 x 2
p2 u0 x k BT
xe te
k BT t
dx u dt
xe dp e
2 mk BT k BT
dx k BT
x 0
x 0 0
p2 u x
0
0 k BT u0
e e
t
e dt
k BT
2 mk BT
dp e k BT
dx dx u
0 0 0
Q3. A cavity contains blackbody radiation in equilibrium at temperature T. The specific heat
per unit volume of the photon gas in the cavity is of the form CV T 3 , where is a
constant. The cavity is expanded to twice its original volume and then allowed to
equilibrate at the same temperature T. The new internal energy per unit volume is
T 4
(a) 4T 4
(b) 2T 4
(c) T 4
(d)
4
Ans: (d)
T 4
Solution: du C v dT T 3dT u
4
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Q4. Consider a system of N non-interacting spins, each of which has classical magnetic
moment of magnitude . The Hamiltonian of this system in an external magnetic field
N
H is i .H , where i is the magnetic moment of the i th spin. The magnetization per
i 1
spin at temperature T is
2H H k B T
(a) (b) coth
k BT k B T H
H H
(c) sinh (d) tanh
k BT k BT
Ans: (b)
2
H cos
cos exp kT
sin d d
Solution: For classical limit M 0 0
H cos
exp kBT sin d d
H k BT
M coth
k BT H
Q5. Consider an ideal Bose gas in three dimensions with the energy-momentum relation
p s with s 0 . The range of s for which this system may undergo a Bose-Einstein
condensation at a non-zero temperature is
(a) 1 s 3 (b) 0 s 2 (c) 0 s 3 (d) 0 s
Ans: (a)
NET/JRF (DEC-2011)
bS 3
Q6. The internal energy E of a system is given by E , where b is a constant and other
VN
symbols have their usual meaning. The temperature of this system is equal to
2
bS 2 3bS 2 bS 3 S
(a) (b) (c) 2 (d)
VN VN V N N
Ans: (b)
E 3bS 2
Solution: TdS dE PdV dE TdS PdV T T
S V VN
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Q7. Consider a Maxwellian distribution of the velocity of the molecules of an ideal gas. Let
Vmp and Vrms denote the most probable velocity and the root mean square velocity,
2kT 3kT V 2
Solution: For Maxwellian distribution Vmp , Vrms mb
m m Vrms 3
Q8. If the number density of a free electron gas in three dimensions is increased eight times,
its Fermi temperature will
(a) increase by a factor of 4 (b) decrease by a factor of 4
(c) increase by a factor of 8 (d) decrease by a factor of 8
Ans: (a)
2
3N 2 3 N
Solution: Fermi energy E F , where is number density and g is degeneracy
4Vg 2m V
2 2
n n
TF1 n1
EF TF K TF TF n
3 2 3
3 1 4 since 8.
V TF2 n2 n2
1
Q9. A system of N non-interacting spin - particles is placed in an external magnetic field H.
2
The behavior of the entropy of the system as a function of energy is given by
(a) S (b) S
E
BH BH B H E B H
S
(c) S (d)
B H E BH B H E
Ans: (a)
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S N U N U N U N U
Solution: ln ln , where H . S is symmetrical
Nk 2N 2 2 N 2N
about E .
Q10. A gas of N non-interacting particles is in thermal equilibrium at temperature T . Each
particle can be in any of the possible non-degenerate states of energy 0, 2 and 4 . The
average energy per particle of the gas, when 1 , is
(a) 2 (b) 3 (c) 2 / 3 (d)
Ans: (a)
0 e o 2 e2 4 e4
Solution: E1 0, E 2 2 , E 3 4 , Z e 0 e 2 e 4 E
e 0 e 2 e 4 1
where 1 .
Q11. A one-dimensional chain consists of a set of N rods each of length a . When stretched
by a load, each rod can align either parallel or perpendicular to the length of the chain.
The energy of a rod is when perpendicular to it. When the chain is in thermal
equilibrium at temperature T , its average length is
(a) Na / 2 (b) Na
(c) Na / 1 e 2 / k BT
(d) Na 1 e 2 / k BT
Ans: (c)
Solution: Let n1 no. of rods are parallel and n2 no. of rods are perpendicular.
Energy of rod when it is perpendicular
Energy of rod when it is parallel is .
e e e
P and P
e e e e e e
n1ae n2 ae Nae Na
Average length n1aP n2 aP
e e e e 1 e 2
Since P P so n2 N , n1 0 .
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Q12. The excitations of a three-dimensional solid are bosonic in nature with their frequency
and wave-number k are related by k 2 in the large wavelength limit. If the chemical
potential is zero, the behavior of the specific heat of the system at low temperature is
proportional to
(a) T 1/ 2 (b) T (c) T 3 / 2 (d) T 3
Ans: (c)
Solution: If dispersion relation is k s ,
At low temperature specific heat T 3/ s
Q13. Gas molecules of mass m are confined in a cylinder of radius R and height L
(with R L ) kept vertically in the Earth’s gravitational field. The average energy of the
gas at low temperatures (such that mgL k BT ) is given by
px2 p 2y p z2 L mgz
Z e e e dpz dx dy e
2 mk BT 2 mk BT 2 mk BT k BT
dpx dp y dz
0
mgL
3 3
mgz
mk T 1 e B
k T
mk T 2 L 2
Z R2 B 2 e k BT
dz Z R 2 B 2
2 0 2 mg
k T
B
ZN Z N ,
ln z 5 Nk BT
E k BT 2 , since mgL k BT
T 2
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Ans: (b)
Solution: We can solve this problem with intuition for example Ak 2
1 3
1
Density of state in 3-dimensional N(ε)
2 2
2
1
Density of state in 2-dimensional N(ε)
0 2
1 1
1
Density of state in 1-dimensional N(ε)
2 2
d
1
Density of state in d-dimensional, where Ak N
s s
Q15. The number of ways in which N identical bosons can be distributed in two energy levels,
is
N N 1 N N 1
(a) N 1 (b) (c) (d) N
2 2
Ans: (a)
Solution: Number of boson N , Number of energy level g
N g 1
So number of ways to distribute N boson into g level is, W cN N 1 since
g 2.
Q16. The free energy of the gas of N particles in a volume V and at a temperature T is
F Nk B T ln a0V k B T
5/ 2
/ N , where a 0 is a constant and k B denotes the Boltzmann
constant. The internal energy of the gas is
3 5
(a) Nk B T (b) Nk B T
2 2
(c) Nk B T ln a 0V k B T
5/ 2
/N
3
2
Nk B T
(d) Nk B T ln a0V / k BT
5/ 2
Ans: (b)
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Solution: F Nk B T ln a0V k B T / N , F U TS , U F TS
5/ 2
F F F
dF SdT PdV S or S U F T
T V T V T V
F Nk B T ln C T a 0Vk B5 / 2
5/ 2
where C
N
F F
Nk B ln CT
5/ 2
Nk B T
C 5 3/ 2
T T Nk B T ln CT
5/ 2 5
Nk B T
T V CT T V
5/ 2
2 2
F 5 F 5
T F Nk B T U F T Nk B T .
T V 2 T V 2
Q17. A system has two normal modes of vibration, with frequencies 1 and 2 21 . What
is the probability that at temperature T , the system has an energy less than 41 ?
[In the following x e 1 and Z is the partition function of the system.]
(a) x 3 / 2 x 2 x 2 / Z (b) x 3 / 2 1 x x 2 / Z
(c) x 3 / 2 1 2 x 2 / Z (d) x 3 / 2 1 x 2 x 2 / Z
Ans: (d)
Solution: There is two normal mode so there is two degree of freedom.
1 1
Energy of harmonic oscillator is E n1 1 n2 2 .
2 2
1 1
E n1 1 n2 21 where n1 0,1,2,3.... and n 2 0,1,2,3....
2 2
31 51
Ground state energy E , first excited state energy E . Second excited state
2 2
71
energy E which is doubly degenerate state so g 2 , other state have more
2
energy than 41 .
3 1 5 1 7 1
PE 41
e
2
e
2
2e
2
x
3
2
1 x 2 x
2
where x e 1 .
Z Z
Q18. Bose condensation occurs in liquid He 4 kept at ambient pressure at 2.17 K . At which
temperature will Bose condensation occur in He 4 in gaseous state, the density of which
is 1000 times smaller than that of liquid He 4 ? (Assume that it is a perfect Bose gas.)
(a) 2.17 mK (b) 21.7 mK (c) 21.7 K (d) 2.17 K
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Ans: (b)
2
N 3
Solution: For bosons T
V
Q19. Consider black body radiation contained in a cavity whose walls are at temperature T .
The radiation is in equilibrium with the walls of the cavity. If the temperature of the walls
is increased to 2T and the radiation is allowed to come to equilibrium at the new
temperature, the entropy of the radiation increases by a factor of
(a) 2 (b) 4 (c) 8 (d) 16
Ans: (c)
8 5 k B4T 4 F 32 5 k B4 3
Solution: For Black Body, energy is given by F V , S VT .
45 2C 3 T V 45 C
3 3
NET/JRF (DEC-2012)
Q20. The entropy of a system, S , is related to the accessible phase space volume by
S k B ln E , N , V where E , N and V are the energy, number of particles and volume
respectively. From this one can conclude that
(a) does not change during evolution to equilibrium
(b) oscillates during evolution to equilibrium
(c) is a maximum at equilibrium
(d) is a minimum at equilibrium
Ans: (c)
Solution: Entropy is maximum at equilibrium.
Q21. Let W be the work done in a quasistatic reversible thermodynamic process. Which of
the following statements about W is correct?
(a) W is a perfect differential if the process is isothermal
(b) W is a perfect differential if the process is adiabatic
(c) W is always a perfect differential
(d) W cannot be a perfect differential
Ans: (b)
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Solution: Work done is perfect differential in adiabatic process.
Q22. The free energy difference between the superconducting and the normal states of a
material is given by F f S f N where is an order parameter and
2 4
2
and are constants s.t. 0 in Normal and 0 in the super conducting state,
while 0 always, minimum value of F is
2 2 3 2 5 2
(a) (b) (c) (d)
2 2 2
Ans: (b)
F 4
Solution: F 2
2 4 3
2 2
2 2 0
3 2
2
2 2
Putting the value, F F
2 2 min
2
Q23. A given quantity of gas is taken from the state A C reversibly, by two P
A
paths, A C directly and A B C as shown in the figure.
During the process A C the work done by the gas is 100 J and the heat
absorbed is 150 J . If during the process A B C the work done by the B C
gas is 30 J , the heat absorbed is V
(a) 20 J (b) 80 J (c) 220 J (d) 280 J
Ans: (b)
Solution: During path AC , dU dQ dW 150 100 50 J
Since, internal energy is point function, so dU will same in all path
In path ABC , dQ dU dW 50 30 80 J .
Q24. Consider a one-dimensional Ising model with N spins, at very low temperatures when
almost all spins are aligned parallel to each other. There will be a few spin flips with each
flip costing an energy 2 J . In a configuration with r spin flips, the energy of the system
is E NJ 2rJ and the number of configuration is N C r ; r varies from 0 to N . The
partition function is
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N N N
J J J
(a) (b) e NJ / k BT
(c) sinh (d) cosh
k BT k B T k B T
Ans: (d)
Solution: Let us consider only three energy levels, E r 2 J 2rJ i.e. E 0 2 J , E1 0 and
E2 2J .
Q2
Ce
2
0
E0
2C1e E1 2C 2 e E2 e 2J
2e 0 e 2 J
e J e J 2
4 4
2
Cr
r 0
2
e J e J
Q2 cosh J 2 cosh J 2 Q N cosh J N .
2
NET/JRF (JUNE-2013)
Q25. Ten grams of ice at 00 C is added to a beaker containing 30 grams of water at 250 C .
What is the final temperature of the system when it comes to thermal equilibrium? (The
specific heat of water is 1 cal / gm / 0 C and latent heat of melting of ice is 80 cal / gm )
mass of the ice. The amount of heat available in water of mass 30 gm at 250 C is
Q m Cv T 30 1 25 750Cal
Since the heat available is less than the heat required to melt the ice therefore ice will not
melt as a result the temperature of the system will be at 00 C only.
Q26. A vessel has two compartments of volume V1 and V2 , containing an ideal gas at pressures
p1 and p 2 , and temperatures T1 and T2 respectively. If the wall separating the
compartments is removed, the resulting equilibrium temperature will be
p1T1 p 2T2 V1T1 V2T2 p1V1 p 2V2
(d) T1T2
1/ 2
(a) (b) (c)
p1 p 2 V1 V2 p1V1 / T1 p 2V2 / T2
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Ans: (c)
p1V1 p2V2
Solution: V V1 V2 , n n1 n2 , U1 U 2 U , n1CvT1 n2CvT2 nCvT ,
T1 T2
p1V1 p2V2
n1T1 n2T2 nT T
p1V1 p2V2
T1 T2
T2
T1
F(v)
v v
T1
T2 T1
F(v)
F(v)
(c) (d)
T2
Ans: (a)
v v
Solution: Area under the F v is conserve and the mean velocity shift towards right for higher
temperature.
Q28. A system of non-interacting spin- 1/ 2 charged particles are placed in an external magnetic
field. At low temperature T , the leading behavior of the excess energy above the ground
state energy, depends on T as: ( c is a constant)
(a) cT (b) cT 3 (c) e c / T (d) c (is independent of T )
Ans: (c)
kTB H H
B
H e e kT
Solution: U B H tanh B B H B H BH
kT e kT e kT
Excess energy from the ground level
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kTB H
B H
B H H
B H
B
e e kT
e kT
e kT
2e kT
B H B H ( B H ) B H 1 B H B H
B H B H
B H B H
e kT e kT e kT e kT
e kT e kT
C
At low temperature, the lower value, U e T
, where C B H .
Q29. Consider a system of two Ising spins S1 and S 2 taking values 1 with interaction energy
given by JS1 S 2 , when it is in thermal equilibrium at temperature T . For large T , the
average energy of the system varies as C / k B T , with C given by
E3 J 1 1 J , E 4 J 1 1 J
Er
J J
Er g r e kT
2 Je
J
2 Je
J
kTJ
e e
J
kT
1 kT 1
kT
J J J
kT kT
U r
J
E
r
J
J
e kT e kT J J
g er
r
kT
2e kT
2e kT
1
kT
1
kT
J2 J
U C J 2 (For large T , 1 )
kT kT
Q30. Consider two different systems each with three identical non-interacting particles. Both
have single particle states with energies 0 ,3 0 and 5 0 , 0 0 . One system is populated
1
by spin fermions and the other by bosons. What is the value of E F E B where E F
2
and EB are the ground state energies of the fermionic and bosonic systems respectively?
E F E B = 5 0 3 0 2 0
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1
Q31. Three identical spin- fermions are to be distributed in two non-degenerate distinct
2
energy levels. The number of ways this can be done is
(a) 8 (b) 4 (c) 3 (d) 2
Ans: (b)
Solution: Total number of degeneracy
g (Number of energy state (n)) (Number of degeneracy due to spin ( 2 s 1 ))
1 1
n 2, s , g 2 (2. 1) 4
2 2
Number of particle, N 3 . So number of ways, g cN 4 c3 4
Q32. Consider the melting transition of ice into water at constant pressure. Which of the
following thermodynamic quantities does not exhibit a discontinuous change across the
phase transition?
(a) Internal energy (b) Helmholtz free energy (c) Gibbs free energy (d) entropy
Ans: (c)
Solution: Ice to water: 1st order phase transition.
So Gibbs free energy is continuous, so it doesn’t exhibit discontinuous change.
Q33. Two different thermodynamic systems are described by the following equations of state:
1 3RN 1 1 5RN 2
and where T 1, 2 , N 1, 2 and U 1, 2 are respectively, the
T 1 2U 1 T 2 2U 2
temperatures, the mole numbers and the internal energies of the two systems, and R is
the gas constant. Let U tot denote the total energy when these two systems are put in
U 1
contact and attain thermal equilibrium. The ratio is
U tot
5 N 2 3N 1 N 1 N 2
(a) (b) (c) (d)
3N 1 5 N 2 3N 1 5 N 2 N 1 N 2 N 1 N 2
Ans: (b)
1 3RN 1 1 5RN 2
Solution: and
T 1 2U 1 T 2 2U 2
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3 5
Now U tot U (1) U 2 RN 1T 1 RN 2T 2
2 2
1
3RN 1T 1
U 1 2 3 N 1T 1
U tot 1 3N 1T 1 5 RN 2T 2 3N 1T 1 5 N 2T 2
2
3N 1
At thermal equilibrium T 1 T 2 , thus
3N 1 5 N 2
Q34. The speed v of the molecules of mass m of an ideal gas obeys Maxwell’s velocity
distribution law at an equilibrium temperature T . Let vx , v y , vz denote the components
2
of the velocity and k B the Boltzmann constant. The average value of vx v y , where
(c) k B T / m (d) k B T / m
2 2
Ans: (b)
Solution: Ideal gas obeys Maxwell velocity distribution law at equilibrium temperature. Then
2
average value of vx v y
v
2
Now x vy 2 v x2 2 v y2 2 v x v y
k BT
v x 0, v y 0 and vx2 v y2 vz2
m
v
2
Then x v y 2 v x2 2 v y2 2 v x v y
v x v y
2
2
k BT
m
k T
2 B 2 2
m
kT
m
Q35. The entropy S of a thermodynamic system as a function of energy S
C
E is given by the following graph. The temperatures of the phases B
A
A, B nd C denoted by T A , TB and TC respectively.
E
Satisfy the following inequalities:
(a) TC TB T A (b) T A TC TB (c) TB TC T A (d) TB T A TC
Ans: (c)
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Solution: Temperatures of phase are: TA , TB , TC
dS 1 S
Since, C
dE T B
dS A
Hence, will be slope, then it will be zero for B - phase
dE E
So TB and in C and A phases, internal energy of C phase is more, so TC T A
Now TB TC T A
Solution: V r
1 2 1
2
Ar A x 2 y 2 z 2 it is harmonic oscillator.
2
3N
1 kT
So its partition function will be z N
N
ln Z N
Internal energy, U kT 2 3 NkT
T
Q37. A Carnot cycle operates as a heat engine between two bodies of equal heat capacity until
their temperatures become equal. If the initial temperatures of the bodies are T1 and T2 ,
respectively and T1 T2 , then their common final temperature is
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NET/JRF (JUNE-2014)
Q38. Which of the graphs below gives the correct qualitative behaviour of the energy density
E r of blackbody radiation of wavelength at two temperatures T1 and T2 T1 T2 ?
(a) (b)
T2
T2
Er
Er
T1
(c) (d)
T2
T2
Er Er
T1
T1
Ans: (c)
Q39. A system can have three energy levels: E 0, . The level E 0 is doubly degenerate,
while the others are non-degenerate. The average energy at inverse temperature is
e e
(a) tanh (d) tanh
(b)
1 e
e
(c) 0
2
Ans: (d)
Solution: E 0, , E 0 doubly degenerate
z gi e Ei 2 e 0 e e
z 2 e e ln z ln 2 e e
Now E
ln z
ln 2 e e
1
2 e e
e e
e 2
e 2
e e
E
tanh
2
2
e 2 e 2 e 2
e 2
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Q40. The free energy F of a system depends on a thermodynamic variable as
F a 2 b 6
with a, b 0 . The value of , when the system is in thermodynamic equilibrium, is
2 F F
F is equilibrium i.e. 0 , now 2a 6b 5
2
F
1/ 4
a a
0 2a 6b 5 4
3b 3b
Q41. For a particular thermodynamic system the entropy S is related to the internal energy U
and volume V by
S cU 3 / 4V 1 / 4
where c is a constant. The Gibbs potential G U TS PV for this system is
3PU cU US
(a) (b) (c) zero (d)
4T 3 4V
Ans: (c)
Solution: S cU 3/ 4V 1/ 4 , dU TdS PdV
S 1 S 1 c 3 1/ 4 1/ 4 4 U 1/ 4
U V T
U V T U V T 4 3c V 1/ 4
1
U S V 5 / 4U 4 S V 5 / 4 1/ 4
P P U
V S c 3 c 3
4 U 1/ 4 S V 5/ 4 1/4 4 1
G U cU V
3/4 1/4
U V U U U 0
3c V 1/ 4
c 3 3 3
Q42. The pressure of a non-relativistic free Fermi gas in three-dimensions depends, at T 0 ,
on the density of fermions n as
(a) n 5 / 3 (b) n1 / 3 (c) n 2 / 3 (d) n 4 / 3
Ans: (a)
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2
Solution: Pressure P nEF , EF n 2 / 3 , at T 0
3
2 2
P n n 2 / 3 n 5 / 3 P n5 / 3
3 3
Q43. The vander Waals’ equation of state for a gas is given by
a
P 2 V b RT
V
where P, V and T represent the pressure, volume and temperature respectively, and a
and b are constant parameters. At the critical point, where all the roots of the above
cubic equation are degenerate, the volume is given by
a a 8a
(a) (b) (c) (d) 3b
9b 27b 2 27bR
Ans: (d)
a P 2 P
Solution: P 2 V b RT , for critical volume 0, 2 0
V V V
a ab
PV Pb 2 RT
V V
P a 2ab 2 P 2a 6ab 2a 6ab
0 P 2 3 0, 0 3 4 0 3 4 Vc 3b
V V V V 2
V V V V
NET/JRF (DEC-2014)
Q44. The pressure P of a fluid is related to its number density by the equation of state
P a b 2
where a and b are constants. If the initial volume of the fluid is V0 , the work done on
the system when it is compressed, so as to increase the number density from an initial
value of 0 to 2 0 is
3a 7 b
(c) 0 0V0 (d) a ln 2 b 0 0V0
2 3
Ans: (d)
n n2 n
Solution: P a b P a b 2
2
V V V
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V2 dV V2 dV n n
W P dV an bn 2 , where V1 , V2
V1 V V1 V 2 0 2 0
Q45. An ideal Bose gas is confined inside a container that is connected to a particle reservoir.
Each particle can occupy a discrete set of single-particle quantum states. If the probability
that a particular quantum state is unoccupied is 0.1 , then the average number of bosons
in that state is
(a) 8 (b) 9 (c) 10 (d) 11
Ans: (b)
Q46. In low density oxygen gas at low temperature, only the translational and rotational modes
of the molecules are excited. The specific heat per molecule of the gas is
1 3 5
(a) kB (b) k B (c) kB (d) kB
2 2 2
Ans: (d)
Solution: Total D.O.F. = 3 transition + 2 rotation i.e. f 5
k B T 5k B T U 5
U f CV kB
2 2 T 2
Q47. When a gas expands adiabatically from volume V1 to V2 by a quasi-static reversible
process, it cools from temperature T1 to T2 . If now the same process is carried out
adiabatically and irreversibly, and T2 is the temperature of the gas when it has
equilibrated, then
V V T2V1
(a) T2 T2 (b) T2 T2 (c) T2 T2 2 1 (d) T2
V2 V2
Ans: (b)
Q48. A random walker takes a step of unit length in the positive direction with probability 2 / 3
and a step of unit length in the negative direction with probability 1 / 3 . The mean
displacement of the walker after n steps is
(a) n / 3 (b) n / 8 (c) 2n / 3 (d) 0
Ans: (a)
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2 1
Solution: P 1 P 1
3 3
2 1 1 n
For one step 1 , for n step
3 3 3 3
Q49. A collection N of non-interacting spins S i , i 1, 2, ....., N , S i 1 is kept in an
external magnetic field B at a temperature T . The Hamiltonian of the system is
B
H B i S i . What should be the minimum value of for which the mean value
k BT
1
Si ?
3
1 1
(a) N ln 2 (b) 2 ln 2 (c) ln 2 (d) N ln 2
2 2
Ans: (c)
B B
e kT
e kT
Solution: P Si 1 B B
, P Si 1 B B
e kT
e kT e kT
e kT
B B
1e kT
e kT
B
Si Si tanh
B B
kT
e kT
e kT
B
For N particle Si N tanh
kT
Si 1 B 1 B 1
According to question, tanh ln 2
N 3 kT 3 kT 2
NET/JRF (JUNE-2015)
Q50. A system of N non-interacting classical particles, each of mass m is in a two
dimensional harmonic potential of the form V r x 2 y 2 where is a positive
1
constant. The canonical partition function of the system at temperature T is :
k BT
N N
2 2m
2N
N
2m 2
(a) (b) (c) (d) 2
2m 2m
Ans. (d)
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Solution: V r x y
2
2
px2 p 2y x 2 y 2
1
z1 2
h
e 2 mkT dpx e 2 mkT dp y e
kT
dx e
kT
dy
2 mkT 2 mkT 1 1 1
z1 2 2
h2 h 2
2 2
kT kT
N
2 m
2
2 m 2
z1 2 kT z N 2 2
2
h h
Q51. A system of N distinguishable particles, each of which can be in one of the two energy
levels 0 and , has a total energy n , where n is an integer. The entropy of the system
is proportional to
N ! N!
(a) N ln n (b) n ln N (d) ln
n ! N n !
(c) ln
n!
Ans. (d)
Solution: No of ways for above configuration is N Cn
N N
W Entropy=k ln
n N n n N n
Q52. The condition for the liquid and vapour phases of a fluid to be in equilibrium is given by
dP Q
the approximate equation 1 (Clausius-Clayperon equation) where vvap is the
dT Tvvap
volume per particle in the vapour phase, and Q1 is the latent heat, which may be taken to
be a constant. If the vapour obeys ideal gas law, which of the following plots is correct?
O O O T O T
T T
Ans. (c)
dP Q RT dP Ql P dP Ql dT C
Solution: l ,
dT Tvap
vap
P
dT RT 2
P
R T 2
ln P
T
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Q53. Consider three Ising spins at the vertices of a triangle which interact with each other with
a ferromagnetic Ising interaction of strength J . The partition function of the system at
1
temperature T is given by :
k BT
(a) 2e3 J 6e J (b) 2e 3 J 6e J
Ans. (b)
Solution: H J S1S 2 S1S3 S 2 S3
S1 S2 S3 E
1 1 1 3J
1 1
1
1 1 1 J
1 1 1
1 11
1 1 1 J
1 1 1
1 1 1 3J
z 2e 3 J 6e J
Q54. A large number N of Brownian particles in one dimension start their diffusive motion
from the origin at time t 0 . The diffusion coefficient is D . The number of particles
crossing a point at a distance L from the origin, per unit time, depends on L and time t
as
L2 4 Dt L2 4 Dt
N NL N
(a) e 4 Dt (b) e L2
(c) e Dt
4
(d) Ne L2
4 Dt 4 Dt 16 Dt 3
Ans. (a)
Solution: From Einstein Smoluchowski theory
dx x2
p x dx exp
4 Dt 4 Dt
N L2
Number of particle passing from point L at origin .exp
4 Dt 4 Dt
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Q55. An ideal Bose gas in d -dimensions obeys the dispersion relation k Ak s , where A
and s are constants. For Bose-Einstein condensation to occur, the occupancy of excited
states
d s
s
Ne c d
0 e
1
where c is a constant, should remain finite even for 0 . This can happen if
d 1 1 d 1 d 1 d
(a) (b) (c) 1 (d) 1
s 4 4 s 2 s 2 s
Ans. (c)
d s
s
Solution: Ne c d
0
e 1
B.E. condensation is possible in 3-D
1
d s 1 d 3
For materlistic particle g 2
s 2 s 2
d s d
For massless particle g 2 2 3
s s
d
In both cases 1
s
NET/JRF (DEC-2015)
Q56. The heat capacity of the interior of a refrigerator is 4.2 kJ / K . The minimum work that
must be done to lower the internal temperature from 18o C to 17o C , when the outside
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N
Solution: For the given system Z N 2sinh F kT ln Z N NkT ln 2sinh
2kT 2kT
2 e 2 kT
e 2 kT
NkT ln e 2 kT 1 e kT NkT ln e 2 kT NkT ln 1 e kT
NkT ln
2
F
kT ln 1 e kT 0 kT 0
N 2 2 2
Q58. The partition function of a system of N Ising spins is Z 1N 2N where 1 and 2 are
functions of temperature, but are independent of N . If 1 2 , the free energy per spin
in the limit N is
(a) k BT ln 1 (b) k BT ln 2 (c) k BT ln 12 (d) k BT ln 1
2
Ans.: (d)
Solution: Z 1N 2N , F kT ln 1N 2N , it is given 1 2
N 2
F kT ln 1N 1 2 , 0
1 1
F
F kT ln 1N NkT ln 1 kT ln 1
N
1
Q59. The Hamiltonian of a system of N non interacting spin particles is H 0 B Siz ,
2 i
B B
2
CV 0 Nk sec2 h 0
kT kT
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0 B
CV 0 B 0 B
2
4
k e kT 2 ln 2
N kT 0 B B
0
2
kT
e kT
e kT
CV 0 B B 16
2 2
4 2 16k 16
ln 2 k ln 2
2
k 0 k
N kT 1
2
kT 25 25 25
2
2
1
Q60. An ensemble of non-interacting spin - particles is in contact with a heat bath at
2
temperature T and is subjected to an external magnetic field. Each particle can be in one
of the two quantum states of energies 0 . If the mean energy per particle is 0 / 2 ,
then the free energy per particle is
(a) 2 0
ln 4 / 3 (b) 0 ln 3 / 2 (c) 2 0 ln 2 (d) 0
ln 2
ln 3 ln 3
Ans.: (a)
0
Solution: For the given system, partition function, Z n 2 N cosh
kT
Mean energy per unit particle 0 0 tanh 0
2 kT
1 e e 1
put 0 tanh
kT 2 e e 2
1 1 2 0
e 2 3 ln 3 0 ln 3 kT
2 kT 2 ln 3
F
kT ln 2 cosh 0 kT ln 2 cosh b 4ac
2
It is given,
N kT
e e
kT ln 2 2
kT ln e e 1 kT ln e 3 1 kT ln e 4
2
0 2
kT ln e ln 4 kT kT ln 4 kT kT ln 4 0 0 ln 4
kT ln 3
3 16
ln 3 2 ln 4 ln 16 ln 3
0 0 0
ln 3 ln 3 ln 3
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4 2
4 4
ln ln ln
F
0
3
2 0
3
2 0
3
N ln 3 ln 3 ln 3
Q61. Which of the following graphs shows the qualitative dependence of the free energy
f h, T of a ferromagnet in an external magnetic field h , and at a fixed temperature
NET/JRF (JUNE-2016)
Q62. The specific heat per molecule of a gas of diatomic molecules at high temperatures is
(a) 8k B (b) 3.5k B (c) 4.5 k B (d) 3k B
Ans: (b)
Solution: For high temperature all number are excited so degree of freedom for diatomic
molecule is 7 .
fk BT 7k T U
Internal energy is , U B , CV 3.5k B
2 2 T V
Q63. When an ideal monoatomic gas is expanded adiabatically from an initial volume V0 to
T
3V0 , its temperature changes from T0 to T . Then the ratio is
T0
2 1
1 1 3 1 3
(a) (b) (c) (d) 3
3 3 3
Ans: (b)
Solution: For adiabatic process PV k , T0V0 1 k
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1 v 1
V 1
T 3V0
1 1
T0V0 T T0 0 T T0
3V0 3
5
For monoatomic gas
3
5 2
1 2/3
1 3 1 3 T 1
T T0 T0
3 3 T0 3
Q64. A box of volume V containing N molecules of an ideal gas, is divided by a wall with a
V
hole into two compartments. If the volume of the smaller compartment is , the
3
variance of the number of particles in it, is
N 2N N
(a) (b) (c) N (d)
3 9 3
Ans: (b)
V 1
Solution: Probability that one particle is in smaller compartment having volume , so p
3 3
There are only two options either particle is in left half or right half, so for one particle
distribution is Bernoulli for Bernoulli’s distribution 2 p 1 p . For N particle
distribution is
1 1 2N
2 Np 1 p N 1 , 2
3 3 9
Q65. A gas of non-relativistic classical particles in one dimension is subjected to a potential
1
V x x (where is a constant). The partition function is
k BT
4m 2m 8m 3m
(a) (b) (c) (d)
h
3 2 2
h
3 2 2
h
3 2 2
3 2 h 2
Ans: (c)
p2 x x
1 x 1
Solution: z e 2 mkT dpx e kT dx 2 mkT e kT dx
1/ 2
h
h
1/ 2 x
2 mkT
z
h
2
e
kT
dx
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x 0 x x
kT kT 2kT
e kT
dx e kT
dx e kT
dx
0
1
2 mkT 2kT 8 m 2
1/ 2
1
z 2 3 2 , put
h h kT
2
Q66. The internal energy E T of a system at a fixed volume is found to depend on the
is
1 2 1 4 4
(a) aT bT (b) 2aT 2 4bT 4 (c) 2aT bT 3 (d) 2aT 2bT 3
2 4 3
Ans: (c)
Solution: From first law of thermodynamics,
TdS dE PdV , dE TdS PdV , it is given dV 0
1
dE TdS dS dE
T
E aT 2 bT 4 dE 2aTdT 4bT 3 dT
4bT 3
dS
1
T
2aTdT 4bT 3dT 2adT 4bT 2 dT 2aT
3
Q67. Consider a gas of Cs atoms at a number density of 1012 atoms/cc. when the typical inter-
particle distance is equal to the thermal de Broglie wavelength of the particles, the
temperature of the gas is nearest to (Take the mass of a Cs atom to be 22.7 1026 kg )
2/3 2
1 h2 1 1012 6.6 1034
T n 2/3
2 mk 2.612 / 3 2 3.14 22.7 1026 1.38 1023 2.612 / 3
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6.6 108 1064 1049
2
6.6 2 107
6.28 22.7 1.38 2.61 6.28 22.7 1.38 2.61
2/3 2/3
NET/JRF (DEC-2016)
Q68. The partition function of a two-level system governed by the Hamiltonian
H
is
(a) 2sinh 2 2
(b) 2 cosh 2 2
2
(d) cosh sinh
1 2 2
2 2
2
Ans. : (b)
Solution: H
Z trace e H e E1 e E2 e 2 2
e
2 2 2 cosh 2 2
Q69. Consider a gas of N classical particles in a two-dimensional square box of side L . If the
total energy of the gas is E , the entropy (apart from an additive constant) is
L2 E LE L E E
(a) Nk B ln (b) Nk B ln (c) 2 Nk B ln (d) L2 k B ln
N N N N
Ans. : (c)
N
1 2 mkTL2 E
Solution: Z N kT
N h2 N
N
1 L2 E 2 m
ZN Assume 1
N N h2
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LE
2
L2 E
ln Z ln N N ln N ln N N N ln
N N
L2 E
F kT ln Z NkT ln N NkT NkT ln
N
E NkT
U F EF L2 E
S Nk Nk ln N Nk Nk ln
T T N
2
L2 E L E L E
Nk ln 2 Nk ln 2 Nk ln
N N N
Q70. Consider a random walk on an infinite two-dimensional triangular lattice, a part of which
is shown in the figure below.
If the probabilities of moving to any of the nearest neighbour sites are equal, what is the
probability that the walker returns to the starting position at the end of exactly three steps?
1 1 1 1
(a) (b) (c) (d)
36 216 18 12
Ans. : (c)
Solution: For walker to return to starting position it must move along an equivalent triangle in
three steps.
For steps one any movement can result in equilateral triangle.
For step two, two out of six options will form equilateral triangle.
For step three, only one out of six options will form equilateral triangle
6 2 1 1
Total probability
6 6 6 18
Q71. An atom has a non-degenerate ground-state and a doubly-degenerate excited state. The
energy difference between the two states is . The specific heat at very low temperatures
1 is given by
(a) k B (b) k B e (c) 2k B e
2
(d) k B
Ans. : (c)
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Solution: Assume energy at ground state is 0 and energy at first excited state is . The partition
function is Z 1 2e
2 e
Energy
1 2e
1 2
2 e 2 2 e kT 2
kT
U kT kT
Specific heat, CV
T V
2
1 2e
kT
1 2e
kT
2 1 2e
kT
2 1 2e
kT
2k e 2k e
kT 1 2e
2 2
1 2e
kT
CV 2k e ,
2
Q72. The electrons in graphene can be thought of as a two-dimensional gas with a linear
energy-momentum relation E p v , where p px , p y and v is a constant. If is the
number of electrons per unit area, the energy per unit area is proportional to
(a) 3/ 2 (b) (c) 1/ 3 (d) 2
Ans. : (a)
Solution: The number of k state in range k to k dk :
2
L
In 2D , it is given by g k dk 2 kdk
2
Since, dispersion relation is E P v kv
2
L EdE L2
g E dE 2 2 2 2
2 v v
2
N
EF2 2 2 v 2 2
L
The average energy at T 0 K is
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EF
E g E dE L2
EF
L2 EF3
Eav 0
N
N 2v 2
0
E 2 dE
3N 2v 2
L 2
2 L2
Eav 2 v
2 2
2 v
2 2 1/ 2
2 v 3/ 2
3N v
2 2
3N
E NEa v 2 E
2 2 v 3/ 2 3/ 2
L2
L 3 L2
NET/JRF (JUNE-2017)
Q73. A thermodynamic function
G T , P, N U TS PV
G G G G
(a) S (b) S (c) V (d)
T N ,P T N ,P P N ,T N P ,T
Ans. : (a)
Solution: G U TS PV
dG dU Tds sdT PdV VdP TdS PdV TdS SdT PdV VdP
dG SdT VdP
G G
S and V
T N , P P N ,T
Q74. A box, separated by a movable wall, has two compartments filled by a monoatomic gas
CP
of . Initially the volumes of the two compartments are equal, but the pressures are
CV
3P0 and P0 respectively. When the wall is allowed to move, the final pressures in the two
compartments become equal. The final pressure is
2 2 31/
(c) 1 31/ P0
1
(a) P0 (b) 3 P0 (d) 1/
P0
3 3 2 1 3
Ans. : (c)
Solution: V1 V2 2V , V2 2V V1 ,
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3PV
0 PV1 , PV0 PV2
P 2V V1
PV
0
P1 T
Q76. In a thermodynamic system in equilibrium, each molecule can exist in three possible
states with probabilities 1/ 2, 1/ 3 and 1/ 6 respectively. The entropy per molecule is
1 2
(a) k B ln 3 (b) k B ln 2 k B ln 3
2 3
2 1 1 1
(c) k B ln 2 k B ln 3 (d) k B ln 2 k B ln 3
3 2 2 6
Ans. : (c)
Solution: S k B Pi ln Pi
i
1
P1 , P2 1/ 3 and P3 1/ 6
2
1
S k B ln1/ 2 1/ 3ln1/ 3 1/ 6 ln1/ 6
2
1 1 1
kb ln1 ln 2 ln1 ln 3 ln1 ln 6
2 3 6
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1 1 1 1 1 1 1 1
k B ln 2 ln 3 ln 2 ln 3 k B ln 2 ln 2 ln 3 ln 3
2 3 6 6 2 6 3 6
3ln 2 ln 2 2ln 3 ln 3 4 ln 2 3ln 3 2 1
S kB kB kB 3 ln 2 2 ln 3
6 6 6 6
Q77. The single particle energy levels of a non-interacting three-dimensional isotropic system,
labelled by momentum k , are proportional to k 3 . The ratio P / of the average pressure
P to the energy density at a fixed temperature, is
(a) 1/ 3 (b) 2 / 3 (c) 1 (d) 3
Ans. : (c)
Solution: E p s , where p is momentum
sE
P , where P is pressure
3V
P s
.
E 3
In problem, E k 3 , so, s 3
3 E E
pressure P P at fixed T .
3V V
Q78. The Hamiltonian for three Ising spins S0 , S1 and S2 , taking values 1 , is
H JS0 S1 S 2
terms of k BT , is
1
1 cosh 2 J
(a) (b) 2 J 1 cosh 2 J
2 sinh 2 J
sinh 2 J
(c) 2 / (d) 2 J
1 cosh 2 J
Ans. : (d)
Solution: H JS0 S1 S 2 J S0 S1 S0 S 2 S0 1 S1 S 2 1
S0 S1 S2 E
1 1 1 2 J
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1 1 1 2J
1 1 1 0
1 1 1 0
1 1 1 0
1 1 1 0
1 1 1 2J
1 1 1 2 J
E1 2 J g1 2
E2 2 J g2 2
E3 0 g3 4
E g e
Ei
U i i
g e i
Ei
0 2 J 2e 2 J 2 J 2e 2 J 4 J e 2 J e 2 J 8 J sinh 2 J
U
4 2 e e
2 J 2 J
4 2e 2e 2 J 2 J
4 4 cosh 2 J
2 J sinh 2 J
U
1 cosh 2 J
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ELECTRONICS AND EXPERIMENTAL METHODS
NET/JRF (JUNE-2011)
C
(high)
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Q3. A time varying signal Vin is fed to an op-amp circuit with output signal Vo as shown in
10 K
the figure below.
The circuit implements a
(a) high pass filter with cutoff frequency 16 Hz 1K
Vo
(b) high pass filter with cutoff frequency 100 Hz V in
1K 10 K
10 K
(c) low pass filter with cutoff frequency 16 Hz 1 F
Ans. : (c)
Solution: Since circuit has R and C combination, its a Low Pass filter and cutoff frequency
1
16 Hz.
2RC
NET/JRF (DEC-2011)
Q4. In the operational amplifier circuit below, the voltage at point A is
5V
1K A
1V
1V
1K
1K 5V
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Q5. A counter consists of four flip-flops connected as shown in the figure:
A0 A1 A2 A3
J Q J Q J Q J Q
CLK
K Q K Q K Q K Q
If the counter is initialized as A0 A1 A2 A3 0110 , the state after the next clock pulse is
0 J Q J Q 0 J Q J Q
CLK
1 K Q K Q 0 1 K Q K Q
1 0 1
Q6. The pins 0, 1, 2 and 3 of part A of a microcontroller are connected with resistors to drive
an LED at various intensities as shown in the figure. For V CC
VCC = 4.2 V and a voltage drop of 1.2 V across the LED,
the range (maximum current) and resolution (step size)
A3
of the drive current are, respectively, 0 . 75 k
A2
(a) 4.0 mA and 1.0 mA C 1 .5 k
A1
(b) 15.0 mA and 1.0 mA 3k
(c) 7.5 mA and 0.5 mA A0
6k
(d) 4.0 mA and 0.5 mA
Ans: (c)
A3 , A2 , A1 , A0
Solution: For Maximum current
0, 0, 0, 0
4.2 1.2 4.2 1.2 4.2 1.2 4.2 1.2
Thus, I max 7.5mA
0.75k 1.5k 3k 6k
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A ,A ,A,A 4.2 1.2
For Step size 3 2 1 0 . Thus I 0 0.5mA
0, 0, 0, 1 6k
Q7. The figure below shows a voltage regulator utilizing a Zener diode of breakdown voltage
5 V and a positive triangular wave input of amplitude 10 V.
500 12
Vi 10
i 8
i(mA)
6
1K 4
2
0
0 1 2 3 4 5 6 7 8
t (s)
For Vi > 5V, the Zener regulates the output voltage by channeling the excess current
through itself. Which of the following waveforms shows the current i passing through the
Zener diode?
(a) 12 (b) 12
10
10
8
i(mA)
8
6
i(mA)
6
4
4
2
2
0
0 1 2 3 4 5 6 7 8 0
0 1 2 3 4 5 6 7 8
t (s) t (s)
(c) 12 (d) 12
10 10
8 8
i(mA)
i(mA)
6 6
4 4
2 2
0 0
0 1 2 3 4 5 6 7 8 0 1 2 3 4 5 6 7 8
t (s) t (s)
Ans: (a)
Solution: When zener is OFF zener current is zero when zener is ON zener current will flow.
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NET/JRF (JUNE-2012)
Q8. In the op-amp circuit shown in the figure below, the input voltage is 1V. The value of the
1K
output V0 is
1K
1K Vo
Vi 1V
1K
Q10. The transistor in the given circuit has hfe = 35Ω and hie = 1000Ω. If the load resistance
RL = 1000Ω, the voltage and current gain are, respectively.
(a) -35 and + 35 VO
RL
(b) 35 and - 35
(c) 35 and – 0.97
VI
Ans. : (a)
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Q11. The output, O, of the given circuit in cases I and II, where
Case I: A, B = 1; C, D = 0; E, F = 1 and G = 0
Case II: A, B = 0; C, D = 0: E, F = 0 and G = 1
are respectively
(a) 1, 0
(b) 0, 1 C
D
(c) 0, 0
(d) 1, 1
Ans. : (d) O
Solution: O AB CD E F G E F
G
NET/JRF (DEC-2012)
Q12. A live music broadcast consists of a radio-wave of frequency 7 MHz, amplitude-
modulated by a microphone output consisting of signals with a maximum frequency of
10 kHz. The spectrum of modulated output will be zero outside the frequency band
(a) 7.00 MHz to 7.01 MHz (b) 6.99 MHz to 7.01 MHz
(c) 6.99 MHz to 7.00 MHz (d) 6.995 MHz to 7.005 MHz
Ans: (b)
Solution: Spectrum consists of f c f m and f c f m .
Q13. In the op-amp circuit shown in the figure, Vi is a sinusoidal input signal of frequency 10
Hz and V0 is the output signal. The magnitude of the gain and the phase shift, respectively,
0.01F
close to the values
(a) 5 2 and / 2
10 K
(b) 5 2 and / 2 1K
Vi
(c) 10 and zero Vo
(d) 10 and π
Ans: (d)
v0 X C RF v
Solution: 0 10
vin R1 R1 R F vin
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Q14. The logic circuit shown in the figure below Implements the Boolean expression
A
HIGH y
(c) 5 V (d) 2 V
Ans: (d)
Solution: 10 v D2 2v D 1 v D 0 v D 2V
Q16. Band-pass and band-reject filters can be implemented by combining a low pass and a
high pass filter in series and in parallel, respectively. If the cut-off frequencies of the low
pass and high pass filters are 0LP and 0HP , respectively, the condition required to
implement the band-pass and band-reject filters are, respectively,
(a) 0HP 0LP and 0HP 0LP (b) 0HP 0LP and 0HP 0LP
(c) 0HP 0LP and 0HP 0LP (d) 0HP 0LP and 0HP 0LP
Ans: (b)
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NET/JRF (JUNE-2013)
Q17. A silicon transistor with built-in voltage 0.7 V is used in the circuit shown, with
V BB 9.7V , R B 300k, VCC 12V and RC 2k . Which of the following figures
RC
RB
V CC
V BB
iC iC
mA
32 35 6 35
(a) 32 (b) 32
Q
30 Q
30
0 9 .7 VCE V 0 12 VCE V
iC iC
mA A
6 35 32 Q 35
(c) Q
32 (d) 32
30 30
0 12 VCE V 0 9 .7 VCE V
Ans: (b)
VBB VBE 9.7 0.7 V 12
Solution: I B 30 A and I C , sat CC 6mA
RB 300 10 3
RC 2 10 3
Q18. If the analog input to an 8-bit successive approximation ADC is increased from 1.0 V to
2.0 V, then the conversion time will
(a) remain unchanged (b) double
(c) decrease to half its original value (d) increase four times
Ans: (a)
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Q19. The input to a lock-in amplifier has the form Vi t Vi sin t i where Vi , , i are the
amplitude, frequency and phase of the input signal respectively. This signal is multiplied
by a reference signal of the same frequency , amplitude Vr and phase r . If the
multiplied signal is fed to a low pass filter of cut-off frequency , then the final output
signal is
1 1
(a) ViVr cos i r (b) ViVr cos i r cos t i r
2 2
1
(c) ViVr sin i r (d) ViVr cos i r cos t i r
2
Ans: (a)
Vi V r
Solution: V Vr sin t r Vi sin t i cos i r cos2 t i r
2
Vi V r
Output of low pass filter= cos i r
2
Q20. Four digital outputs V , P, T and H monitor the speed v , tyre pressure p , temperature t
and relative humidity h of a car. These outputs switch from 0 to 1 when the values of the
parameters exceed 85 km/hr, 2 bar, 40 0 C and 50%, respectively. A logic circuit that is
used to switch ON a lamp at the output E is shown below.
Which of the following condition will not switch the lamp ON?
(a) v 85km / hr , p 2 bar , t 40 0 C , h 50%
V
(b) v 85km / hr , p 2 bar , t 40 C , h 50%
0
P
(c) v 85km / hr , p 2 bar , t 40 0 C , h 50% E
H
Ans: (d)
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JRF/NET-(DEC-2013)
Q21. Consider the op-amp circuit shown in the figure.
If the input is a sinusoidal wave Vi 5 sin 1000t , then 1 F
vo 10 3 1 1 5 5 2
3 vo sin t sin t
vin 2 10 2 2 2
Q22. If one of the inputs of a J-K flip flop is high and the other is low, then the outputs Q and
Q
(a) oscillate between low and high in race around condition
(b) toggle and the circuit acts like a T flip flop
(c) are opposite to the inputs
(d) follow the inputs and the circuit acts like an R S flip flop
Ans: (d)
Q23. A sample of Si has electron and hole mobilities of 0.13 and 0.05 m 2 /V- s respectively at
300 K. It is doped with P and Al with doping densities of 1.5 10 21 / m 3 and
2.5 10 21 / m 3 respectively. The conductivity of the doped Si sample at 300 K is
(a) 8 1 m 1 (b) 32 1 m 1 (c) 20.8 1 m 1 (d) 83.2 1 m 1
Ans: (a)
Solution: Resulting doped crystal is p-type and p p 2.5 1.5 10 21 / m 3 1 10 21 / m 3
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Q24. Two identical Zener diodes are placed back to back in series and are connected to a
variable DC power supply. The best representation of the I-V characteristics of the circuit
is
(a) I (b) I
V V
(c) I (d) I
V V
Ans: (d)
Q25. A 4-variable switching function is given by f 5, 7, 8, 10, 13, 15 d 0, 1, 2 , where
Ans: (d) CD CD CD CD
AB BD
AB 1 1 BD
AB 1 1
AB 1 1
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NET/JRF (JUNE-2014)
Q26. The inner shield of a triaxial conductor is driven by an (ideal) op-amp follower circuit as
shown. The effective capacitance between the signal-carrying conductor and ground is
Signal
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Q29. For the logic circuit shown in the below
X
B
A A
(c) (d)
B B
X X
C C
Ans: (d)
Solution:
A A
A.B A B AC
B X
B
C
C
ABC
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NET/JRF (DEC-2014)
Q30. Consider the amplifier circuit comprising of the two op-amps A1 and A2 as shown in the
figure. 1M
R
10 K
r
A1 V0
A2
10 V
F
F V where V is the uncertainty in the measurement of volume.
V
F 1 2 / 3
F V 1/3 V
V 3
1 1 1 1
F V 2.7 2.7 2.7 F 0.09
3 30 3 900 3 9.7
2/3 2/3 1/ 3
3V
Q33. Consider a Low Pass (LP) and a High Pass (HP) filter with cut-off frequencies f LP and
f HP , respectively, connected in series or in parallel configurations as shown in the
Figures A and B below.
fHP
(A) Input L Output (B) Input Output
fHP fLP
L
Which of the following statements is correct? fLP
(a) For f HP f LP , A acts as a Band Pass filter and B acts as a band Reject filter
(b) For f HP f LP , A stops the signal from passing through and B passes the signal
without filtering
(c) For f HP f LP , A acts as a Band Pass filter and B passes the signal without filtering
(d) For f HP f LP , A passes the signal without filtering and B acts as a Band Reject filter
Ans: (c)
Q34. The power density of sunlight incident on a solar cell is 100 mW / cm 2 . Its short circuit
current density is 30 mA / cm 2 and the open circuit voltage is 0.7 V . If the fill factor of
the solar cell decreases from 0.8 to 0.5 then the percentage efficiency will decrease from
(a) 42.0 to 26.2 (b) 24.0 to 16.8 (c) 21.0 to 10.5 (d) 16.8 to 10.5
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Ans: (d)
Solution: The efficiency of a solar cell is determined as the fraction of incident power which is
converted to electricity and is defined as
Voc I sc FF
and Pmax Voc I sc FF
Pin
where Voc is the open circuit voltage, I sc is the short circuit current density , FF is the
Fill factor, Pin is the input power and is the efficiency of the solar cell.
NET/JRF (JUNE-2015)
Q35. The concentration of electrons, n and holes p , for an intrinsic semiconductor at a
3
E
temperature T can be expressed as n p AT 2 exp g , where Eg is the band
2 k BT
3
gap and A is a constant. If the mobility of both types of carrier is proportional to T 2
,
then the log of the conductivity is a linear function of T 1 , with slope
Eg Eg Eg Eg
(a) (b) (c) (d)
2k B kB 2k B kB
Ans. (c)
3
Eg 23 Eg
Solution: i ni e e p T 2 exp T i C exp
2 k BT 2k B T
Eg Eg
ln i ln C slope is
2 k BT 2k B
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Pa 4
Q36. The viscosity of a liquid is given by Poiseuille’s formula . Assume that
8lV
l and V can be measured very accurately, but the pressure P has an rms error of 1% and
the radius a has an independent rms error of 3% . The rms error of the viscosity is
closest to
(a) 2% (b) 4% (c) 12% (d) 13%
Ans. (c)
Solution: k pa 4
2
p2 a2 a 4 p2 4 Pa 3 a2
2 2 2
n
P a
2
p
2 2
n a
n 100 p 100 16 a 100 1 16 3 1 144 145
2 2
n
n 100 12%
Q37. Consider the circuits shown in figures (a) and (b) below
2K 1K
10 K 10 K
10V 10V
10.7V 5V
(a) (b)
If the transistors in Figures (a) and (b) have current gain dc of 100 and 10 respectively,
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Thus VCB VC VB 10 2 100 0.7 ve
decreases by 150 per 0 C upon heating. The output voltage of the circuit at 300 C is
T
1V
1K
Vout
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NET/JRF (DEC-2015)
Q39. If the reverse bias voltage of a silicon varactor is increased by a factor of 2 , the
corresponding transition capacitance
(a) increases by a factor of 2 (b) increases by a factor of 2
(c) decreases by a factor of 2 (d) decreases by a factor of 2
Ans.: (c)
1 CT V C V 1
Solution: CT T CT CT
V CT V CT 2V 2
Q40. If the parameters y and x are related by y log x , then the circuit that can be used to
(a) y (b) y
Vo Vo
(c) (d)
y y
Vo Vo
Ans.: (c)
Solution: (1) Integrator
(2) Logarithmic Ampere V0 log y
(4) Differentiator
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Q41. The state diagram corresponding to the following circuit is
x D A
y
CLOCK
Flip Flop
00, 01,10 01,11
(a) 11 00 (b) 00,10 00,10
0 1 0 1
01,10,11 01,11
0 1 0 1
0 0 0 1 1
0 0 1 0 0
0 1 0 1 1
0 1 1 0 0
1 0 0 1 1
1 0 1 0 0
1 1 0 0 0
1 1 1 1 1
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Q42. A sinusoidal signal of peak to peak amplitude 1V and unknown time period is input to
the following circuit for 5 second’s duration. If the counter measures a value 3E 8 H in
NET/JRF (JUNE-2016)
Q43. The dependence of current I on the voltage V of a certain device is given by
2
V
I I 0 1
V0
where I 0 and V0 are constants. In an experiment the current I is measured as the voltage
V applied across the device is increased. The parameters V0 and I 0 can be graphically
determined as
(a) the slope and the y -intercept of the I V 2 graph
(b) the negative of the ratio of the y -intercept and the slope, and the y -intercept of the
I V 2 graph
(c) the slope and the y -intercept of the I V graph
(d) the negative of the ratio of the y -intercept and the slope, and the y -intercept of the
I V graph
Ans: (d)
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V
2
V I0 I
Solution: I I 0 1 I I 0 1 I V I0
V0 V0 V0
I0 I0
Slope V0
V0 I0 V
V0
Intercept on y -axis I 0
Q44. In the schematic figure given below, assume that the propagation delay of each logic gate
is t . 5 V
gate
The propagation delay of the circuit will be maximum when the logic inputs A and B
make the transition
(a) 0,1 1,1 (b) 1,1 0,1
(c) 0, 0 1,1 (d) 0, 0 0,1
Ans: (d)
Solution:
Input Output
A B NOT OR AND OR
0 1 0 0 0 0
3t
1 1 0 1 1 1
1 1 0 1 1 1
3t
0 1 0 0 0 0
0 0 1 1 1 1
t
1 1 0 1 1 1
0 0 1 1 1 1
4t
0 1 0 0 0 0
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Q45. Given the input voltage Vi , which of the following waveforms correctly represents the
0.5 10 K
5K
Vi 0 Vi
t
V0
0.5V
5K 10 K
0.5
0
3.0 3.0
(a) 2.5 (b) 2.5
2.0 t V0 2.0
V0
1.5 1.5
1.0 1.0 t
0.5 0.5
0.0 0.0
0 0
3.0 3.0
(c) 2.5 (d) 2.5
2.0 2.0 t
V0 1.5 V0 1.5
1.0 t 1.0
0.5 0.5
0.0 0.00
0
Ans: (b)
10 10 10
Solution: V0 1 0.5 Vi V0 1 2Vi
5 15 5
When Vi 0 V0 1V , when Vi 0.1V V0 0.8 V , when Vi 0.5V V0 0V
Q46. The decay constants f p of the heavy pseudo-scalar mesons, in the heavy quark limit, are
a
related to their masses m p by the relation f p , where a is an empirical parameter
mp
Ans. : (c)
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Solution: a f p m p
1/ 2
2 2
a 2 a 2 a a f
2
f p m p m1/p 2 and p1
a f p
m p f p m p
2m p2
1
f 2
f
2
2
m
2
f 2
mp
2
2
a2 m p 2f p m2 p a a p
p
p
a
4m p p
a fp
2
2m p
f p 2m p
2 2
fp 15 2 3
mp 160 2 4
6.9 10 and 1.56 10
fp 180 2m p 2 6400
MeV
1/ 2
a 180 80 MeV
3/ 2 1/ 2 3/ 2
6.9 103 1.56 104 180 80 7 103
a 1204 MeV
3/ 2
NET/JRF (DEC-2016)
Q47. Which of the following circuits implements the Boolean function
F A, B, C 1, 2, 4, 6 ?
C I0 C I0
(a) I1 4 1 (b) I1 4 1
I MUX
2
F I MUX
2
F
I 3 S1 S0 I 3 S1 S0
A B A B
C I0 0 I0
(c) I1 4 1 (d) I1 4 1
1
I MUX
2
F I MUX
2
F
I 3 S1 S0 C I 3 S1 S0
A B A B
Ans. : (b)
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Solution:
A B C F
0 0 0 0
0 0 1 1 F C
0 1 0 1
0 1 1 0 F C
1 0 0 1
1 0 1 0 F C
1 1 0 1
1 1 1 0 F C
aV
Q48. The I V characteristics of a device can be expressed as I I s exp 1 , where T
T
is the temperature and a and I s are constants independent of T and V . Which one of
1 1
2
0 3 3
1 2 3 0 1 2
aV / T aV / T
2 4
1 3
(c) (d)
2
0 1
log I
log I
1 0
1
2
2
3 3 3
0 1 2 0 1 2 3
aV / T aV / T
Ans. : (d)
av
Solution: Let x For large x ; I I s e x log e I log e Is x log e I x
T
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Q49. The active medium in a blue LED (light emitting diode) is a Gax In1 x N alloy. The band
gaps of GaN and InN are 3.5 eV and 1.5 eV respectively. If the band gap of Gax In1 x N
varies approximately linearly with x , the value of x required for the emission of blue
light of wavelength 400 nm is (take hc 1200 eV -nm )
(a) 0.95 (b) 0.75 (c) 0.50 (d) 0.33
Ans. : (b)
Solution: EgGaN 3.5eV and EgInN 1.5eV E eV
Vi R1 RL
R2
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Q51. Two sinusoidal signals are sent to an analog multiplier of scale factor 1V 1 followed by a
low pass filter (LPF).
V1 5cos 100t
LPF
Multiplier fC 5Hz
Vout
V2 20 cos 100t / 3
1
50 cos 200t
3 2
1
After pass L.P.F. v0 50 25V
2
Q52. The resistance of a sample is measured as a function of temperature, and the data are
shown below.
T 0C 2 4 6 8
The slope of R vs T graph, using a linear least-squares fit to the data, will be
6 4 2 8
(a) 0
(b) 0
(c) 0
(d) 0
C C C C
Ans. : (b)
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NET/JRF (JUNE-2017)
Q53. In the n -channel JFET shown in figure below, Vi 2V , C 10 pF , VDD 16 V and
RD 2k . VDD
RD
D C
VO
Vi
G
S
If the drain D - source S saturation current I DSS is 10 mA and the pinch-off voltage VP is
1
Q54. The gain of the circuit given below is .
RC
C
R V
Vin
a Vout
b
V
ground
The modification in the circuit required to introduce a dc feedback is to add a resistor
(a) between a and b
(b) between positive terminal of the op-amp and ground
(c) in series with C
(d) parallel to C
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Ans. : (d)
Q55. A 2 4 decoder with an enable input can function as a
(a) 4 1 multiplexer (b) 1 4 demultiplexer
(c) 4 2 encoder (d) 4 2 priority encoder
Ans. : (b)
Q56. The experimentally measured values of the variables x and y are 2.00 0.05 and
3.00 0.02 respectively. What is the error in the calculated value of z 3 y 2 x from the
measurements?
(a) 0.12 (b) 0.05 (c) 0.03 (d) 0.07
Ans. : (a)
Solution: z 3 y 2 x
2 2
z z
y2 x2 9 y2 4 x2 0.12
2
z
y x
Q57. Let I 0 be the saturation current, the ideality factor and vF and vR the forward and
reverse potentials respectively, for a diode. The ratio RR / RF of its reverse and forward
vR qv vF qv
(a) exp F (b) exp F
vF k BT vR k BT
vR qv vF qv
(c) exp F (d) exp F
vF k BT vR k BT
Ans. : (a)
KT
Solution: I I 0 eV /VT 1 , VT
q
RR VR / I R VR I F
RF VF / I F VF I R
RR VR I 0 eVF /VT VR qV
exp F
RF VF I0 VF KT
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Q58. In the figures below, X and Y are one bit inputs. The circuit which corresponds to a one
bit comparator is
X
X Y
(a) Y
X Y
X Y
X
X Y
(b)
X Y
Y X Y
X
(c) X Y
X Y
Y X Y
X
X Y
(d)
Y
X Y
Y X Y
Ans. : (c)
Solution: (a) 01 XY , 02 XY , 03 0
(b). 01 XY , 02 XY , 03 Y
(d) 01 XY , 02 X Y , 03 XY
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Q59. Both the data points and a linear fit to the current vs voltage of a resistor are shown in the
graph below. 1
I (amps)
0 V volts 25
If the error in the slope is 1.255 103 1 , then the value of resistance estimated from the
graph is
(a) 0.04 0.8 (b) 25.0 0.8
Ans. : (b)
I max I min 1 0 1
Solution: Slope m (let)
Vmax Vmin 25 0 25
V 1 R 1
I mV R 25 where 2
R m m m
2
R 2 1 2
Error in R is R2 m 4 m R m
4 2
m m
Q60. In the following operational amplifier circuit Cin 10 nF , Rin 20 k , RF 200 k and
RF
CF 100 pF .
CF
Ri Ci
Vin
Vout
Vo RF / J cF RF 1 RF j ci
Vin jci Ri 1 / jci jcF RF 1 1 j Ri ci
Vo ci RF
, 2 f
Vin 1 cF RF 1 Ri ci
2 2
V0
2 16 103 10 109 200 103
Vln 1 4 2 16 103 200 103 100 1012 1 4 2 16 103 20 103 10 109
2 2 2 2 2 2
64
4.96
20.12 20.12
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ATOMIC AND MOLECULAR PHYSICS
NET/JRF (JUNE-2011)
Q1. Consider the energy level diagram (as shown in the figure below) of a typical three level
ruby laser system with 1.6 1019 Chromium ions per cubic centimeter. All the atoms
excited by the 0.4 μm radiation decay rapidly to level E2 , which has a lifetime = 3 ms.
E3
0.4 m
E2
0.7 m
E1
(c) N 2 t
Rt 2
1 e
t /
(d) N2 (t) = R t
Ans: (b)
Solution: N 2 t R 1 et /
B. The minimum pump power required (per cubic centimeter) to bring the system to
transparency, i.e. zero gain, is
(a) 1.52 kW (b) 2.64 kW (c) 0.76 kW (d) 1.32 kW
Ans: (c)
Solution: The Minimum Power required to achieve zero gain is
N hv N hc 1.6 1019 6.6 10 34 3 10 8
P 754 W cm 3
2 2 2 6
0.7 10 3 10 3
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NET/JRF (DEC-2011)
Q2. Given that the ground state energy of the hydrogen atom is –13.6 eV, the ground state
energy of positronium (which is a bound state of an electron and a positron) is
(a) + 6.8 eV (b) – 6.8 eV (c) – 13.6 eV (d) – 27.2 eV
Ans: (b)
13.6
Solution: The energy expression for Positronium atom is E n eV
2n 2
13.6
For n = 1, E1 eV 6.8eV , E1 6.8 eV
2
Q3. A laser operating at 500 nm is used to excite a molecule. If the Stokes line is observed at
770 cm-1, the approximate positions of the Stokes and the anti-Stokes lines are
(a) 481.5 nm and 520 nm (b) 481.5 nm and 500 nm
(c) 500 nm and 520 nm (d) 500 nm and 600 nm
Ans:
Solution: Given 0 500 nm 5 105 cm , v stoke 770 cm 1 0 20,000cm 1
the electron and proton spins, respectively, the splitting between the 3 S1 and 1 S 0 state is
a 3 a a 3 3 2
H1 2 2 2 for 3 S1 and H 2 0 2 a for 1 S 0
2 2 4 2 2 4
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1 3
The splitting between 3 S1 and 1 S 0 is H H 1 H 2 a 2 a 2
4 4
Q5. The ratio of intensities of the D1 and D2 lines of sodium at high temperature is
(a) 1:1 (b) 2:3 (c) 1:3 (d) 1:2
Ans: (d)
Solution: The electronic transition for D2 and D1 line is
3
2 1
I D2 2 J 2 1 2 4 2
D2 : 2P3/ 2 2S1/ 2 , D1 : 2P1/ 2 2S1/ 2
I D1 2 J 1 1 1 2 1
2 1
2
Q6. An atom of mass M can be excited to a state of mass M by photon capture. The
frequency of a photon which can cause this transition is
c 2 c 2 2 c 2 c 2
(a) (b) (c) (d) 2M
2h h 2Mh 2Mh
Ans: (d)
Solution: The conservation law of energy and momentum give
Mc 2 h M c 4 p 2 c 2
2
1/ 2
and
h
c
p
c 2 c 2
2Mc h 2Mc 1
2
4
1 2M .
2M h 2M 2Mh
NET/JRF (JUNE-2012)
12
Q7. The first absorption spectrum of C16O is at 3.842 cm-1 while that of 13
C16O is at
3.673 cm-1. The ratio of their moments of inertia is
(a) 1.851 (b) 1.286 (c) 1.046 (d) 1.038
Ans: (c)
Solution: For 12 C 16O : 2 B1 3.842 cm 1 B1 1.921 cm 1
For 13
2 B 2 3.673 cm 1
C 16 O : B 2 1.8365 cm 1
h I 2 B1 1.921
Where, B 2 1.046
8 IC I1 B2 1.8365
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Q8. The spin-orbit interaction in an atom is given by H aL S , where L and S denote the
orbital and spin angular momenta, respectively, of the electron. The splitting between the
levels 2P3/2 and 2P1/2 is
3 2 1 2 5 2
(a) a (b) a (c) 3a 2 2 (d) a
2 2 2
Ans. : (a)
Solution: Given H aL S where J L S .
1
a
J 2 L2 S 2 2 L S L S J 2 L2 S 2 H J 2 L2 S 2
2 2
1 3
For 2P3/ 2 : S which gives S 2 S S 1 2 2
2 4
1 3
For 2P1/ 2 : S which gives S 2 S S 1 2 2
2 4
L = 1 which gives L2 LL 1 2 2 2
1 3 a 3 3
J which gives J 2 J J 1 2 2 H 2 2 2 a 2
2 4 2 4 4
a 2 3
H H1 H 2 a 2 a 2
2 2
Q9. The spectral line corresponding to an atomic transition from J = 1 to J = 0 states splits in
a magnetic field of 0.1 Tesla into three components separated by 1.6 10-3 Ǻ. If the zero
field spectral line corresponds to 1849 Ǻ, what is the g-factor corresponding to the J = 1
hc
state? (You may use 2 10 4 cm)
0
(a) 2 (b) 3/2 (c) 1 (d) 1/2
Ans: (c)
Solution: The Zeeman splitting is E gM J B B g B B for MJ = +1
2
Given, Zeeman splitting separations, 1.6 10 3 Å
c
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c 3 10 8
2 1.6 10 1 0.1404 1010
1849 10 10 2
E 6.625 10 34 0.1404 1010
g 1.00
B B 9.27 10 24 0.1
NET/JRF (DEC-2012)
Q10. Consider the energy level diagram shown below, which corresponds to the molecular
nitrogen laser. 2
R 21
1
1
0
20 -3 -1
If the pump rate R is 10 atoms cm s and the decay routes are as shown with
21 20 ns and 1 1s , the equilibrium populations of states 2 and 1 are, respectively,
(a) 1014 cm-3 and 2 1012 cm-3 (b) 21012 cm-3 and 1014 cm-3.
(c) 21012 cm-3 and 2 10 6 cm-3 (d) zero and 1020 cm-3
Ans: (b)
dN 2 N dN1 N 2 N 1
Solution: R 2 and .
dt 21 dt 21 1
dN 2 dN1
Under equilibrium condition 0
dt dt
N 2 21 R 1020 20 109 2 1012 cm3
1 N 2 10 6 2 1012 cm 3
N1 1014 cm 3
21 20 10 9
Q11. Consider a hydrogen atom undergoing a 2 P 1S transition. The lifetime tsp of the 2P
state for spontaneous emission is 1.6 ns and the energy difference between the levels is
10.2 eV. Assuming that the refractive index of the medium n0 = 1, the ratio of Einstein
coefficients for stimulated and spontaneous emission B21 / A21 is given by
(a) 0.683 × 1012 m3J-1s-1 (b) 0.146 × 10-12 Jsm-3 .
(c) 6.83 × 1012 m3J-1s-1 (d) 1.463 × 10-12 Jsm-3 .
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Ans: (a)
B21 2 c 3 2 2 c 3
Solution: n 0 1 , E 10.2 eV and 0.67 1012 .
A21 3 n03 E 3 n03
Q12. Consider a He-Ne laser cavity consisting of two mirrors of reflectivities R1 = 1 and
R2 = 0.98. The mirrors are separated by a distance d = 20 cm and the medium in between
has a refractive index n0 = 1 and absorption coefficient α = 0. The values of the
separation between the modes and the width p of each mode of the laser cavity are:
Ans: (c)
c
Solution: Mode separation 750 MHz
2dn0
1 2n 0 d
Width of each mode p where t c .
2t c 1
c ln 2d
R1 R2 e
Note: In this question, there is no need to calculate p . Since in the given options there
is only one option with 750 MHz and i.e., option (c). You can calculate p without
NET/JRF (JUNE-2013)
Q13. A muon
from cosmic rays is trapped by a proton to form a hydrogen-like atom.
Given that a muon is approximately 200 times heavier than an electron, the longest
wavelength of the spectral line (in the analogue of the Lyman series) of such an atom will
be
o o o o
(a) 5.62 A (b) 6.67 A (c) 3.75 A (d) 13.3 A
Ans: (b)
m m p
Solution: In case of muonic atom, the reduce mass is m ' 180me
m m p
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m E
'
E
E n' 21 180 21 where, E1 13.6eV
me n n
E
2
6.3 10 16 eV sec 2
Q15. The electronic energy levels in a hydrogen atom are given by E n 13.6 / n 2 eV. If a
selective excitation to the n 100 level is to be made using a laser, the maximum
allowed frequency line-width of the laser is
(a) 6.5 MHz (b) 6.5 GHz (c) 6.5 Hz (d) 6.5 kHz
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Ans: (b)
2 13.6 2 13.6
Solution: E n 13.6 / n 2 E n 3
eV h 1.6 10 19 6.5 GHz .
n n3
Q16. Consider the laser resonator cavity shown in the figure.
If I1 is the intensity of the radiation at R1 1 R2 R
l
mirror M 1 and is the gain coefficient of the
medium between the mirrors, then the energy
x
density of photons in the plane P at a
P
distance x from M 1 is M1 M2
(a) I1 / c e x (b) I1 / c e x
energy level of the 2 P state, when the effects of spin are neglected up to second order in
a , is
3 2 4
(a) 0 (b) 2a 2 a 2 4 (c) 2a 2 (d) a 2 a
2
Ans: (c)
Solution: For 2 P state, L=1
The shift in the energy due to perturbation V pert aL2 is
strong magnetic field (when L S coupling can be neglected) the ground state energy
level will split into
(a) 3 levels (b) 4 levels (c) 5 levels (d) 6 levels
Ans: (c)
Solution: In extremely strong magnetic field coupling between L-S breaks down. J is no longer a
valid quantum number.
The ground state energy level will have 2S 1 2 L 1 6 states, but 2 states are
NET/JRF (JUNE-2014)
Q20. A spectral line due to a transition from an electronic state p to an s state splits into three
Zeeman lines in the presence of a strong magnetic field. At intermediate field strengths
the number of spectral lines is
(a) 10 (b) 3 (c) 6 (d) 9
Ans: (a)
Solution: For p state: l 1, s 1/ 2 : j 1/ 2 & 3 / 2 .
This gives two spectral terms 2P3/2 & 2P1/2
For s state: l= 0, s = 1/2: j = 1/2 : This gives spectral terms 2S1/2
The terms 2P3/2 and 2S1/2 corresponding to J = 3/2 & J = 1/2 will break into 2J+1 Zeeman
levels, which is 4 and 2 respectively.
For 2P3/2 Mj = - 3/2 -1/2 +1/2 +3/2
For 2S1/2 Mj = -1/2 +1/2
The selection rule is ∆MJ = 0, ± 1 (MJ = 0 → MJ = 0 If ∆J = 0)
∆MJ = 0 gives π component, ∆MJ = ±1 gives σ component
Number of π component = 2, Numbers of + components = 2
Number of - components = 2
The terms 2P1/2 and 2S1/2 corresponding to J = 1/2 & J = 1/2 will break into 2J+1 Zeeman
levels, which is 2 & 2 respectively.
For 2P1/2 Mj = -1/2 +1/2, For 2S1/2 Mj = -1/2 +1/2
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The selection rule is ∆MJ = 0, ± 1 (MJ = 0 → MJ = 0 If ∆J = 0)
∆MJ = 0 gives π component, ∆MJ = ±1 gives σ component
Number of π component = 2, Numbers of + components = 1
Number of - components = 1
Thus, total number of Zeeman component = 10
Q21. A double slit interference experiment uses a laser emitting light of two adjacent
frequencies v1 and v 2 v1 v 2 . The minimum path difference between the interfering
beams for which the interference pattern disappears is
c c c c
(a) (b) (c) (d)
v 2 v1 v 2 v1 2v 2 v1 2v 2 v1
Ans: (c)
Solution: The condition of maximum intensity for interfering laser beam is: d sin n
1
The condition of dark intensity for interfering laser beam is: d sin n
2
For interference pattern to vanish, the minimum path difference should be /2
c c 2
The spectral bandwidth of laser is defined as, v 2
v
For two closely spaced line of wavelength 1 and 2
c 2 1 c c
1 2
v 1 2 1 1
c c 2 1
2 1 2 1
Since, for interference pattern to vanish for two closely spaced line of wavelength 1 and
c c
2, the minimum path difference should be =
2 2v 2 2 1
Q22. How much does the total angular momentum quantum number J change in the transition
of Cr 3d 6 atom as it ionize to Cr 2 3d 4 ?
(a) Increases by 2 (b) Decreases by 2 (c) Decreases by 4 (d) Does not change
Ans: (c)
Solution: In Cr 3d 6 state
M L 2 1 0 1 2
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1
In this configuration, S 4 2 and L 2
2
This is the case of more than half filled subshell, thus state with highest J value will
have the lowest energy.
The highest J - value is J L S y
Now in Cr 2 3d 4
M L 2 1 0 1 2
1
In this configuration, S 4 2 and L 2
2
Since this is the case of less than half filled subshell, thus, state with lowest J value will
have the lowest energy.
The lowest J - value is J L 5 2 2 0 .
Thus the ground state spectral term for this configuration is the J - value decreases from
J 4 to J 0 .
Thus correct answer is option (c).
NET/JRF (DEC-2014)
Q23. An atomic transition 1 P 1S in a magnetic field 1 Tesla shows Zeeman splitting. Given
that the Bohr magneton B 9.27 10 24 J / T , and the wavelength corresponding to the
transition is 250 nm, the separation in the Zeeman spectral lines is approximately
(a) 0.01 nm (b) 0.1 nm (c) 1.0 nm (d) 10 nm
Ans: (a)
Solution: This is the case of Normal Zeeman effect. The Zeeman separation in terms of
B
frequency, B , where B is Bohr magneton
h
2 2 B B
In terms of wavelength,
c c h
12
m 0.003 nm
3 10 m / s 6.625 10 Js
8 34
None of the answer is matching correctly. But best suitable answer is option (a)
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Q24. If the leading anharmonic correction to the energy of n th vibrational level of a diatomic
2
1
molecule is xe n with xe 0.001 , the total number of energy levels possible
2
is approximately
(a) 500 (b) 1000 (c) 250 (d) 750
Ans: (a)
2
1 1
Solution: The energy of anharmonic oscillator is Ev v xe v
2 2
where v 0, 1, 2, .....vmax is vibrational quantum number v vmax
dEv 1
Now, 0 2 xe vmax 0
dv v vmax
2 E
1 1 1 1 1
1 2 xe vmax vmax 500
2 2 xe 2 2 xe 2 0.001
Q25. The effective spin-spin interaction between the electron spin S e and the proton spin S p
in the ground state of the Hydrogen atom is given by H aS e S p . As a result of this
The coupling between Se and S p gives net resultant spin angular momentum
1
S Se S p , S 2 Se2 S p2 2Se S p Se S p S 2 Se2 S p2
2 Sp
S
H
2
a 2
S Se2 S p2
Se
where S 2 S S 1 2 , Se2 Se Se 1 2 , S p2 S p S p 1 2 F 1
1 2
a
1 1 4
Since Se and S p S 0, 1 12 s1/2
2 2 3
a 2
4
F 0
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a 3 3 3
For S 0 (singlet state), H1 0 2 a 2
2 4 4 4
a 3 2 3 2 1 2
For S 1 (Triplet state), H 2 2 a
2 4 4 4
1 2 3 2
H H 2 H1 a a H a 2
4 4
NET/JRF (JUNE-2015)
Q26. Of the following term symbols of the np 2 atomic configurations, 1 S0 ,3 P0 ,3 P1 ,3 P2 and
1
D2 which is the ground state?
(a) 3 P0 (b) 1 S0 (c) 3 P2 (d) 3 P1
Ans. (a)
Solution: According to Hund’s rules
(i) State with highest multiplicity has lowest energy
(ii) State with same multiplicity, the state with highest L will have lowest energy
(iii) State with same multiplicity and L value. The state with lowest J has lowest energy
(only if subshell is less than half filled) from the given states 1 S0 ,3 P0 ,3 P1 ,3 P21 D2
the transitions in which both the initial and final states are restricted to v 1 and j 2
and subject to the selection rules v 1 and j 1 . Then the largest allowed energy
of transition is
(a) 3B (b) B (c) 4 B (d) 2 B
Ans. (c)
1
Solution: E v BJ J 1
2
For vibrational transition with v 1 and rotational transition with J we get
E Einitial Efinal 2 B J 1
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where J is lowest quantum number
According to question in rotational states, two transition is possible, one for
J 0 J 1 and second for J 1 J 2
But, second transition will have photon of higher energy
For J 1 J 2
E 2 B 1 1 4 B
NET/JRF (DEC-2015)
Q28. The LS configurations of the ground state of 12Mg , 13
Al , 17Cl and 18
Ar are respectively,
Ground state: 1 S0
13
Al :1s 2 2s 2 2 p 6 3s 2 3 p1
the terms are 2 p1/ 2 and 2 p3/ 2 . Since its less than half filled. Thus
Ground state: 1 S0
Q29. For a two level system, the population of atoms in the upper and lower levels are 3 1018
and 0.7 1018 , respectively. If the coefficient of stimulated emission is
3.0 105 m3 / W -s 3 and the energy density is 9.0 J / m3 -Hz , the rate of stimulated
emission will be
(a) 6.3 1016 s 1 (b) 4.1 1016 s 1 (c) 2.7 1016 s 1 (d) 1.8 1016 s 1
Ans.: None of the answer is matching.
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Solution: The rate of stimulated emission is
dN 2
N 2 B21 u w
dt
m3 m3
where N 2 3 1018 , B21 3 105 3 10 5
W s3 J s2
J s
and u w 9.0 J / m3 Hz 9.0
m3
dN 2 m3 J s
3 1018 3 105 9 3 8.11024 s 1
dt J s 2
m
Q30. The first ionization potential of K is 4.34 eV , the electron affinity of Cl is 3.82 eV
and the equilibrium separation of KCl is 0.3 nm . Then energy required to dissociate a
KCl molecule into a K and a Cl atom is
(a) 8.62 eV (b) 8.16 eV (c) 4.28 eV (d) 4.14 eV
Ans.: (c)
Solution: Energy required to dissociate KCl is KCl K Cl
2
1 q1q2 9 Nm
2 1.6 1019 c
V 9 10 2
7.7 1019 J 4.79 eV
4 0 r12 c 0.3 10 m9
The band dissociation energy is the energy required to dissociate a molecule into its
component atom KCl K Cl
To find the energy required to dissociate KCl into K and Cl , we must add an electron
to the K ion, which releases the atomic potassium ionization energy. Remove one
electron from Cl ion which requires the atomic chlorine electron affinity energy
Given ionization energy of K Eie 4.34 eV
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NET/JRF (JUNE-2016)
Q31. The ground state electronic configuration of 22 Ti is Ar 3d 2 4s 2 . Which state, in the
1 1
s1 , s2 S 0,1
2 2
Now, S 0, L 4 J 4, 1G4
S 0, L 2 J 2, 1 D2
S 0, L 0 J 0, 1S0
S 1, L 3 J 4,3, 2, 3
F4 , 3F3 , 3F2
S 1, L 1 J 2,1, 0, 3 P2 , 3 P1 ,3 P0
Thus 1 F3 is not possible spectroscopic term of Ar 3d 2 4 S 2
Q32. In a normal Zeeman Effect experiment using a magnetic field of strength 0.3 T , the
splitting between the components of a 660 nm spectral line is
(a) 12 pm (b) 10 pm (c) 8 pm (d) 6 pm
Ans: (d)
2 eB 660 10
2
9
1.6 1019 0.3
Solution: 6.09 1012 m 6 pm
c 4 m 3 108 4 9.110 31
Q33. The separation between the energy levels of a two-level atom is 2 eV . Suppose that
4 1020 atoms are in the ground state and 7 1020 atoms are pumped into the excited state
just before lasing starts. How much energy will be released in a single laser pulse?
(a) 24.6 J (b) 22.4 J (c) 98 J (d) 48 J
Ans: (d)
Solution: N 2 N1 3 1020
N N1 3 1020
Energy of laser pulse, E 2 h 2 1.6 1019 J E 48 J
2 2
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NET/JRF (DEC-2016)
Q34. In the L S coupling scheme, the terms arising from two non-equivalent p -electrons are
S 0, L 1 J 1 ; 1 P1 or 1 P
S 0, L 2 J 2 ; 1 D2 or 1 D
S 1, L 0 J 1 ; 3 S1 or 3 S
S 1, L 1 J 0,1, 2 ; 3 P0,1,2 or 3 P
S 1, L 2 J 1, 2,3 ; 3 D1,2,3 or 3 D
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Q36. A two level system in a thermal (black body) environment can decay from the excited
state by both spontaneous and thermally stimulated emission. At room temperature
300 K , the frequency below which thermal emission dominates over spontaneous
emission is nearest to
(a) 1013 Hz (b) 108 Hz (c) 105 Hz (d) 1011 Hz
Ans. : (d)
Solution: At thermal equilibrium, the ratio of the number of spontaneous to stimulated emission
is given by
A21 1.054 1034 J .S
e kT 1 ; where 2.551014 sec
B21 u 23
kT 1.38 10 J / K 300 K
3000 10 10 3
2.7 1020
7.7 108 sec 80n sec
3.5 1013
Q39. If the binding energies of the electron in the K and L shells of silver atom are 25.4 keV
and 3.34 keV , respectively, then the kinetic energy of the Auger electron will be
approximately
(a) 22 keV (b) 9.3keV (c) 10.5 keV (d) 18.7 keV
Ans. : (d)
Solution: K .E. Auger electron is K .E EK EL EL
EK 2 EL Auger electorn
e
25.4 2 3.34 keV
Vacuum level
18.7keV
L
h
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CONDENSED MATTER PHYSICS
NET/JRF (JUNE-2011)
Q1. A narrow beam of X-rays with wavelength 1.5 Ǻ is reflected from an ionic crystal with an
fcc lattice structure with a density of 3.32 gcm-3. The molecular weight is 108 AMU
(1AMU = 1.66 × 10-24 g)
A. The lattice constant is
(a) 6.00 Ǻ (b) 4.56 Ǻ (c) 4.00 Ǻ (d) 2.56 Ǻ
Ans: (a)
Solution: Given ne f f 4 , M = 108 kg, 3.32gm cm -3 = 3320 kgm-3,
ne f f M 4 108
a3 6.00 10 30 m 3 6.00 10 10 6.00 A 0
NA 6.023 10 26 3320
B. The sine of the angle corresponding to (111) reflection is
3 3 1 1
(a) (b) (c) (d)
4 8 4 8
Ans: (b)
Solution: According to Bragg’s law
a a
2d sin , sin where d for (111) plane
2d h2 k 2 l 2 3
3 3 1.5 A0 3 3 3
sin .
2a 2 6A 0
2 6 2 8
Q2. A flux quantum (fluxoid) is approximately equal to 210-7 gauss-cm2. A type II
superconductor is placed in a small magnetic field, which is then slowly increased till the
field starts penetrating the superconductor. The strength of the field at this point is
2
105 gauss.
A. The penetrating depth of this superconductor is
(a) 100 Ǻ (b) 10Ǻ (c) 1000Ǻ (d) 314Ǻ
Ans: (a)
Solution: Given Fluxoid 0 2 10 7 gauss –cm2
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2
First Critical field H c1 10 5 gauss
The relation between first critical field and penetration depth is
0 0 2.10 7
H c1 2
10 12 cm 2 10 6 cm 100 A 0
2
H c1 2 10 5
B. The applied field is further increased till superconductivity is completely destroyed.
8
The strength of the field is now 105 gauss. The correlation length of the
superconductor is
(a) 20 Ǻ (b) 200 Ǻ (c) 628 Ǻ (d) 2000 Ǻ
Ans: None of the options is matched.
8
Solution: Given second critical field H c 2 10 5 gauss. The relation between second critical
0
field and correlation length is H c 2 2.
0 2 10 7 1 1 100
2
10 12 cm 2 106 cm 1010 m 50 A 0
H c 2 8
10 5 4
2 2
Q3. The two dimensional lattice of graphene is an arrangement of Carbon atoms forming a
honeycomb lattice of lattice spacing a, as shown below. The Carbon atoms occupy the
vertices.
1
c1
1
c2
1
3 2
1
a1
1 1
(a) 2a 2 (b) a d1 b1
2
1
a2 b2
a
2 3 3 2
(c) 6 3a (d) a
2
Ans: (d)
Solution: Primitive lattice vectors are b1 and b2
b1 3a cos 300 iˆ 3a cos 600 ˆj
2
3
a 3iˆ ˆj
3 3 2
b2
2
3
a
3i ˆj , A b2 b1
2
a
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(B). The Bravais lattice for this array is a
(a) rectangular lattice with basis vectors d1 and d 2
(b) rectangular lattice with basis vectors c1 and c2
(c) hexagonal lattice with basis vectors a1 and a2
(d) hexagonal lattice with basis vectors b1 and b2
Ans: (c)
Solution: The Bravaiss lattice for this array is the Hexagonal lattice with basis vectors a1 and a 2
NET/JRF (DEC-2011)
Q4. The potential of a diatomic molecule as a function of the distance r between the atoms is
a b
given by V r 6
12 . The value of the potential at equilibrium separation between
r r
the atoms is:
(a) 4a 2 / b (b) 2a 2 / b (c) a 2 / 2b (d) a 2 / 4b
Ans: (d)
a b dV r
Solution: Given V r 6
12 . At equilibrium radius, 0
r r dr r r0
dV r 6a 12b r 13 12b 2b 2b
7 13 0 07 r06
dr r0 r0 r0 6a a a
a b a2 a2 a2
The value of potential at equilibrium is V r0 .
r06 r012 2b 4b 4b
Q5. If the number density of a free electron gas in three dimensions is increased eight times,
its Fermi temperature will
(a) increase by a factor of 4 (b) decrease by a factor of 4
(c) increase by a factor of 8 (d) decrease by a factor of 8
Ans: (a)
Solution: The relation between Fermi energy and electron density is E F
2
2m
3 2 n
2/3
.
E F'
2
2m
3 2 8n 2/3
4EF .
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Q6. The excitations of a three-dimensional solid are bosonic in nature with their frequency ω
and wave-number k are related by ω k2 in the large wavelength limit. If the chemical
potential is zero, the behaviour of the specific heat of the system at low temperature is
proportional to
(a) T1/2 (b) T (c) T3/2 (d) T3
Ans: (c)
Solution: If the dispersion relation is k s in large wavelength. Then the specific heat is
C v T 3 / s . Given k 2 C v T 3 / 2
NET/JRF (JUNE-2012)
Q7. Consider a system of non-interacting particles in d dimensional obeying the dispersion
relation Ak s , where ε is the energy, k is the wavevector, s is an integer and A is
constant. The density of states, N(ε), is proportional to
s d d s
1 1 1 1
(a) d
(b) s
(c) s
(d) d
Ans: (b)
Q8. The experimentally measured transmission spectra of metal, insulator and semiconductor
thin films are shown in the figure. It can be inferred that I, II and III correspond,
respectively, to
Transmission (%)
Q9. The energy required to create a lattice vacancy in a crystal is equal to 1 eV. The ratio of
the number densities of vacancies n(1200 K)/n(300 K) when the crystal is at equilibrium
at 1200 K and 300 K, respectively, is approximately
(a) exp 30 (b) exp 15 (c) exp15 (d) exp30
Ans: (d)
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Solution: The equation for number density of vacancies n Ne E / 2 k BT where E: Energy required
to form vacancies, N: density of lattice sites
E 1 1
n1200K
E 1 1 E 1
n e E / 2 k BT1
2k T T
1 E / 2 k BT2 e B 2 1 , e 2 k B 300 1200 e 2 k B 400 e 30
n2 e n300 K
Q10. The dispersion relation of phonons in a solid is given by
2 k 02 3 cos k x a cos k y a cos k z a
The velocity of the phonons at large wavelength is
(a) 0 a / 3 (b) 0 a (c) 3 0 a (d) 0 a / 2
Ans: (d)
Solution: For large , k x a, k y a, k z a are small.
kya
2 2
k x2 a 2 k z2 a 2 02 a 2 2
2 k 02 3 1
1
2
2
1
2
2
k x k y2 k z2
2a2 a d 0 a
2 k 0 k 2 0 k v g .
2 2 dk 2
NET/JRF (DEC-2012)
Q11. A magnetic field sensor based on the Hall Effect is to be fabricated by implanting As into
a Si film of thickness 1 μm. The specifications require a magnetic field sensitivity of
500 mV/Tesla at an excitation current of 1 mA. The implantation dose is to be adjusted
such that the average carrier density, after activation, is
(a) 1.25 × 1026 m-3 (b) 1.25 × 1022 m-3
(c) 4.1 × 1021 m-3 (d) 4.1 × 1020 m-3
Ans: (b)
IB 10 3 1 V
Solution: n 6 19
3
1.25 10 22 m 3 where H 500 10 3 V / T .
teVH 10 1.6 10 500 10 B
Q12. In a band structure calculation, the dispersion relation for electrons is found to be
k cos k x a cos k y a cos k z a ,
where β is a constant and a is the lattice constant. The effective mass at the boundary of
the first Brillouin zone is
2 2 4 2 2 2
(a) (b) (c) (d)
5 a 2 5 a 2 2 a 2 3 a 2
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Ans: (d)
2
Solution: k cos k x a cos k y a cos k z a , Effective mass m
d 2 k
2
d k
Brilliouin zone boundary is at k x ,ky , kz .
a a a
d 2 2
Hence 2 k 3 a 2 m .
d k , , 3 a 2
a a a
Q13. The radius of the Fermi sphere of free electrons in a monovalent metal with an fcc
structure, in which the volume of the unit cell is a3, is
1/ 3 1/ 3 1/ 3
12 2 3 2 2 1
(a) 3 (b) 3 (c) 3 (d)
a a a a
Ans: (a)
1/ 3
3 2 N 2 2 k F2
2/3
Solution: Radius of Fermi sphere is k F , EF 3 2
n
V 2m 2m
1/ 3
N 4 12 2
For fcc solid k F 3 .
V a3 a
NET/JRF (JUNE-2013)
Q14. Using the frequency-dependent Drude formula, what is the effective kinetic inductance of
a metallic wire that is to be used as a transmission line? [In the following, the electron
mass is m , density of electrons is n , and the length and cross-sectional area of the wire
and A respectively.]
(a) mA / ne 2 (b) zero (c) m / ne 2 A
(d) m A / ne 2 2
Ans: (c)
Q15. The phonon dispersion for the following one-dimensional diatomic lattice with masses
M 1 and M 2 (as shown in the figure)
K
M1 M2 M1 M2
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is given by
1 1 4M 1 M 2 qa
2 q K 1 1 sin 2
M 1 M 2 M 1 M 2 2
2
where a is the lattice parameter and K is the spring constant. The velocity of sound is
K M 1 M 2 K
(a) a (b) a
2M 1 M 2 2M 1 M 2
K M 1 M 2 KM 1 M 2
(c) a (d) a
2M 1 M 2
3
M 1M 2
Ans: (b)
Solution: For small value of q (i.e. long wavelength approximation limit).
qa qa
We have sin
2 2
1 1 4M 1M 2 qa
2 q 1 1 sin 2
M1 M 2
2
M1 M 2 2
1 4M 1 M 2 qa
2
1
q
2
1 1 2
M
1 M 2
M 1 M 2 2
1 1 1 4 M 1 M 2 q 2 a 2
2 q 1 1
2 M M 2 4
M
1 M 2 1 2
1 1 M 1 M 2 q 2 a 2
q
2
1 1
M M 2 2
M
1 M 2 1 2
1 1 M 1 M 2 q 2 a 2
For Acoustical branch: 2 q 1 1
M M 2 2
M
1 M 2 1 2
M M2 M 1 M 2 q 2 a 2 a 2
2 q 1
M M 2 2 2M M q
2
M 1M 2 1 2 1 2
q aq
2M 1 M 2
Velocity of sound is v g a
q 2M 1 M 2
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Q16. The electron dispersion relation for a one-dimensional metal is given by
ka 1 2
k 2 0 sin 2 sin ka
2 6
where k is the momentum, a is the lattice constant, 0 is a constant having dimensions of
energy and ka . If the average number of electrons per atom in the conduction band
g E dE 1 L2
E E.
dE u 2 2
NET/JRF (DEC-2013)
Q18. The physical phenomenon that cannot be used for memory storage applications is
(a) large variation in magnetoresistance as a function of applied magnetic field
(b) variation in magnetization of a ferromagnet as a function of applied magnetic field
(c) variation in polarization of a ferroelectric as a function of applied electric field
(d) variation in resistance of a metal as a function of applied electric field
Ans: (d)
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Q19. The energy of an electron in a band as a function of its wave vector k is given
by E k E 0 B cos k x a cos k y a cos k z a , where E0 , B and a are constants. The
2 2 2
1 2 1
E0 B 3 a 2 k x k x k x E0 3B Ba 2 k 2
2 2
2
2
Effective mass of the electron is m * 2
d E / dk 2 Ba 2
Q20. A DC voltage V is applied across a Josephson junction between two superconductors
with a phase difference 0 . If I 0 and k are constants that depend on the properties of the
junction, the current flowing through it has the form
2eVt 2eVt
(a) I 0 sin 0 (b) kV sin 0
(c) kV sin 0 (d) I 0 sin 0 kV
Ans: (a)
Q21. A uniform linear monoatomic chain is modeled by a spring-mass system of masses m
separated by nearest neighbour distance a and spring constant m 02 . The dispersion
relation for this system is
ka ka
(a) k 2 0 1 cos (b) k 2 0 sin 2
2 2
ka ka
(c) k 2 0 sin (d) k 2 0 tan
2 2
Ans: (c)
Solution: The dispersion relation for uniform linear mono-atomic chain of atoms is
ka
k 2 0 sin
2
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NET/JRF (JUNE-2014)
Q22. The pressure of a nonrelativistic free Fermi gas in three-dimensions depends, at T 0 ,
on the density of fermions n as
(a) n 5 / 3 (b) n1 / 3 (c) n 2 / 3 (d) n 4 / 3
Ans: (a)
Solution: The Fermi energy in three dimension is defined as
2/3
2 3 2 N
EF
2m V
2
2m
3 2 n 2/3
2
5
2
nEF n
5
2
2m
3 n
2 2/3 2 2
5 2m
3 2 n 5 / 3
2/3
Q23. Consider an electron in b.c.c. lattice with lattice constant a . A single particle
wavefunction that satisfies the Bloch theorem will have the form f r exp ik .r , with
f r being
2 2 2
(a) 1 cos x y z cos x y z cos x y z
a a a
2 2 2
(b) 1 cos x y cos y z cos z x
a a a
(c) 1 cos x y cos y z cos z x
a a a
(d) 1 cos x y z cos x y z cos x y z
a a a
Ans: (b)
Solution: The primitive translational vector for BCC is
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a
a
a
a ' iˆ ˆj kˆ , b ' iˆ ˆj kˆ , c ' iˆ ˆj kˆ
2 2 2
Bloch function defined as
k r u k r e ik .r f r e ik .r
Here f r is atomic wavefunction, which has the periodicity of the lattice i.e.
u k r a u k r
Given Bloch function
2 2 2
f (r ) 1 cos x y cos y z cos z x
a a a
2 a a 2 a a 2 a a
f (r a ' ) 1 cos x y cos y z cos z x
a 2 2 a 2 2 a 2 2
2 2 2
f (r a ' ) 1 cos x y cos y z 2 cos z x
a a a
2 2 2
f (r a ' ) 1 cos x y cos y z cos z x f (r )
a a a
f (r a ' ) f (r )
Similarly,
f (r b ' ) f (r ) and f ( r c ' ) f ( r )
Other functions do not satisfy the periodicity
Q24. The dispersion relation for electrons in an f.c.c. crystal is given, in the tight binding
approximation, by
kxa kya kya k a k a k a
k 4 0 cos cos cos cos z cos z cos x
2 2 2 2 2 2
where a is the lattice constant and 0 is a constant with the dimension of energy. The x -
component of the velocity of the electron at , 0, 0 is
a
(a) 2 0 a / (b) 2 0 a / (c) 4 0 a / (d) 4 0 a /
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Ans: (d)
Solution: Group velocity of electron in dispersive medium is expressed as
1 d 1 d ˆ d ˆ d
v i j kˆ v x iˆ v y ˆj v z kˆ
dk dk x dk y dk z
kx a kya ka k a ka kya kya k a
sin cos cos z sin x iˆ cos x sin sin cos z ˆj
2 a 2 2 2 2 2 2 2 2
v 0
sin k z a cos y cos k x a sin k z a kˆ
k a
2 2 2 2
At , 0, 0
a
2 a
v 0 sin cos 0 cos 0sin iˆ cos sin 0 sin 0cos 0 ˆj cos 0sin 0 sin 0cos kˆ
2 2 2 2
4 a
v 0 iˆ 0 ˆj 0kˆ 0iˆ 0 ˆj 0kˆ vx iˆ v y ˆj vz kˆ
4 a
vx 0 , v y 0, vz 0
4 0 a
The x - component of velocity is vx
NET/JRF (DEC-2014)
Q25. When laser light of wavelength falls on a metal scale with 1 mm engravings at a
grazing angle of incidence, it is diffracted to form a vertical chain of diffraction spots on
a screen kept perpendicular to the scale. If the wavelength of the laser is increased by 200
nm, the angle of the first-order diffraction spot changes from 5 0 to
(a) 6.60 0 (b) 5.14 0 (c) 5.018 0 (d) 5.210
Ans: (c)
Solution: The condition of maxima peak in grating is
b sin m ; m 0,1, 2,3,....
where b is the width of slit or width of engraving, whereas ‘ m ’ is the order of
diffraction and is the angle of diffraction
For 1st order diffraction: b sin (i)
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When wavelength of incident light increased to 200 nm , let’s assume the 1st order
Ans: (b)
Solution: This question can only be solved by solving each option by assuming r1 r2 and
comparing result with the packing fraction of simple cubic which is .
6
r 3 r 3 1 3 1 3
Option (a): 1
2
r1 r2 r1 r2 2 2 4
2 r13 r23 2 2r 3 2 1
Option (b):
3 r1 r2 3 3 8r 3 3 4 6
r13 r23 2r 3 1
Option (c): 3
r1 r2
3
8r 4
r13 r23 2r 3
Option (d):
2 r1 r2 3 2 8r 3
8
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Q27. Consider two crystalline solids, one of which has a simple cubic structure, and the other
has a tetragonal structure. The effective spring constant between atoms in the c -direction
is half the effective spring constant between atoms in the a and b directions. At low
temperatures, the behaviour of the lattice contribution to the specific heat will depend as
a function of temperature T as
(a) T 2 for the tetragonal solid, but as T 3 for the simple cubic solid
(b) T for the tetragonal solid, and as T 3 for the simple cubic solid
(c) T for both solids
(d) T 3 for both solids
Ans: (d)
Solution: The specific heat of solid in three dimensions is proportional to T 3 and it is
independent of crystal structure.
In 3D : CV T 3
In 2D : CV T 2
In 1D : CV T
Q28. A superconducting ring carries a steady current in the presence of a magnetic field B
normal to the plane of the ring. Identify the incorrect statement.
(a) The flux passing through the superconductor is quantized in units of hc / e
(b) The current and the magnetic field in the superconductor are time independent.
(c) The current density j and B are related by the equation j 2 B 0 , where
is a constant
(d) The superconductor shows an energy gap which is proportional to the transition
temperature of the superconductor
Ans: (a)
Solution: The flux quantization in superconducting ring is no
hc h
where o in CGS units and o in MKS units.
2e 2e
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NET/JRF (JUNE-2015)
Q29. X -ray of wavelength a is reflected from the 111 plane of a simple cubic lattice. If
a 3 3
sin sin 1
2d a 2 2 3
2
3
Q30. The critical magnetic fields of a superconductor at temperatures 4 K and 8 K are
11 mA / m and 5.5 mA / m respectively. The transition temperature is approximately
(a) 8.4 K (b) 10.6 K (c) 12.9 K (d) 15.0 K
Ans. (b)
Solution: The relation between critical field and critical temperature is
T 2
H C T H 0 1
TC
T 2 T 2
Thus we get H C T1 H 0 1 , H C T2 H 0 1 2
1
TC TC
H C T1 2
2
T
1 1 T2 T12
H C T1 C 2 2 8 4
2 2
T H T
C T TC 10.6
H C T2 T
2 C
H C T1 2 1
1 2 1
H C T2
TC
where T1 4 k , T2 8 k , H C T1 11 mA / m and H C T2 5.5 mA / m
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Q31. The low-energy electronic excitations in a two-dimensional sheet of grapheme is given
by E k vk , where v is the velocity of the excitations. The density of states is
proportional to
3 1
(a) E (b) E 2 (c) E 2 (d) E 2
Ans. (a)
Solution: The number of k - states in range k and k dk in two dimension is
2
L
g k dk 2 kdk
2
2 2
L E dE L 2
E k dE dk g E dE 2 EdE
2 2
2
T 104 K
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NET/JRF (DEC-2015)
Q33. The first order diffraction peak of a crystalline solid occurs at a scattering angle of 300
when the diffraction pattern is recorded using an x-ray beam of wavelength 0.15 nm . If
the error in measurements of the wavelength and the angle are 0.01 nm and 10
respectively, then the error in calculating the inter-planar spacing will approximately be
(a) 1.1102 nm (b) 1.3 104 nm (c) 2.5 102 nm (d) 2.0 103 nm
Ans.: (a)
d 1 d cos
Solution: Bragg’s Law for n 1, 2d sin d ,
2sin 2sin 2sin 2
Error in d can be calculated as
2 2 2
d d 2 1 2 cos 2
2
2
2sin 2sin
d 2
2 2 2
d2 1 2sin 2 cos 2sin 2
2
d 4sin
2
2sin sin
1
2
2
2 2 2 2
d
2 d d
d 2
2
tan tan
1.5 1010
d 1.5 1010 m
2sin 2sin 30o
1
2 2 1
0.11010 2
3
2 2
d 1.5 10 0.0734 0.111010 1.11011 m 1.1102 nm
10
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Q34. The dispersion relation of electrons in a 3-dimensional lattice in the tight binding
approximation is given by,
k cos k x a cos k y a cos k z a
where a is the lattice constant and , , are constants with dimension of energy. The
effective mass tensor at the corner of the first Brillouin zone , , is
a a a
1/ 0 0 1/ 0 0
2 2
(a) 2 0 1/ 0 (b) 2 0 1/ 0
a a
0 0 1/ 0 0 1/
1/ 0 0 1/ 0 0
2 2
(c) 2 0 1/ 0 (d) 2 0 1/ 0
a a
0 0 1/ 0 0 1/
Ans.: (c)
Solution: The effective mass as a tensor quantity can be written as
m*xx m*xy m*xz
* 2
mij* m*yx m*yy m yz where mij
*
2E
m*zx m*zy m*zz
ki k j
since k cos k x a cos k y a cos k z a
2 2 2 2
m*xx , m*yy
2 a 2 cos k x a 2 a 2 cos k y a
2
k x k x k y
2 2
m 2
*
, other terms are zero
a 2 cos k z a
zz
2
k z
1/ 0 0
* 2 2 2 2
Now, at , , ; mxx , m yy
*
, mzz 2 mij 2 0 1/
* *
0
a a a a 2
a 2
a a
0 0 1/
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Q35. A thin metal film of dimension 2 mm 2 mm contains 4 1012 electrons. The magnitude
of the Fermi wavevector of the system, in the free electron approximation, is
(a) 2 107 cm 1 (b) 2 107 cm 1 (c) 107 cm 1 (d) 2 107 cm 1
Ans.: (b)
Solution: This is the case of two dimensional metal box. The Fermi wave vector of electron in
2 D is
1
1
N 2
k F 2 n 2 2 2 ; L2 2mm 2mm 4 102 cm 2 ,
L
1
4 10 12 1
2 10 cm 2 107 cm 1
2
k F 2 2 2
14 2 2
4 10 cm
Q36. For an electron moving through a one-dimensional periodic lattice of periodicity a ,
which of the following corresponds to an energy eigenfunction consistent with Bloch’s
theorem?
x x x 2 x
(a) x A exp i cos (b) x A exp i cos
a 2a a a
2 x 2 x x x
(c) x A exp i i cosh (d) x A exp i i
a a a 2a
Ans.: (b)
Solution: According to block theorem, x a x
2 x 2 x
x a A exp i x a cos x a A exp i cos 2
a a a a
2 x x 2 x
A exp i x a cos A exp i cos
a a a a
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NET/JRF (JUNE-2016)
Q37. Consider electrons in graphene, which is a planar monoatomic layer of carbon atoms. If
the dispersion relation of the electrons is taken to be k ck (where c is constant)
over the entire k -space, then the Fermi energy F depends on the number density of
electrons as
1 2 1
(a) F 2 (b) F (c) F 3 (d) F 3
Ans: (a)
Solution: In 2 D , density of state is
L d
g k dk 2 kdk , where ck k and dk
2 c c
2
L d L2
g d 2 . d
2 c c 2 c 2
Now, number electrons at T 0 K is
F L2 F L2 2 N
N g d d F2 4 c 2 2 4 c 2
0 2 c 2 0 4 c 2 F
L
F 4 c 2 1/ 2 F 1/ 2
Q38. Suppose the frequency of phonons in a one-dimensional chain of atoms is proportional to
the wave vector. If n is the number density of atoms and c is the speed of the phonons,
then the Debye frequency is
cn
(a) 2 cn (b) 2 cn (c) 3 cn (d)
2
Ans: (d)
Solution: Given k ck ( c is velocity of phonon)
L d L
Now g d d
d / dk c
D L D L
Also N g d d N D
0 c 0 c
N N cn
D c c n, n f D . Best answer is (d).
L L 2
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Q39. The band energy of an electron in a crystal for a particular k -direction has the form
k A B cos 2ka , where A and B are positive constants and 0 ka . The
electron has a hole-like behaviour over the following range of k :
3 3
(a) ka (b) ka (c) 0 ka (d) ka
4 4 2 4 2 4
Ans: (a)
d d 2
Solution: k A B cos 2ka , 2 Ba sin 2ka , 2
4 Ba 2 cos 2ka
dk dk
2 2
Effective mass m* 2
d / dk 2 2
4 Ba cos 2ka
Effective mass of electron me* and effective mass of holes mh* are opposite in sign i.e.,
m *
h m . *
e
Now, in the range 0 ka , m* is positive
4
3
While in the range ka , m* is negative
4 4
3
Thus, electron has hole like behaviour in the region ka
4 4
NET/JRF (DEC-2016)
Q40. Consider a hexagonal lattice with basis vectors as shown in the figure below.
y
x
a2 a
1
a
If the lattice spacing is a 1 , the reciprocal lattice vectors are
4 2 2 4 2 2
(a) ,0, , (b) ,0, ,
3 3 3 3 3 3
4 2 2 2 2
(c) 0, , , (d) , , 2 ,
3 3 3 3 3
Ans. : (a)
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Solution: From the figure, we can write
a1
3a
2
3 xˆ yˆ , a2 3a yˆ , a3 azˆ (let us assume)
Now V a1. a2 a3
3a
2
3 xˆ yˆ . 3ayˆ azˆ
3a
2
3 xˆ yˆ . 3a 2 xˆ 3 3 3
2
a
Also, a3 a1 azˆ
3a
2
3x y
ˆ ˆ
2
3a 2
3 yˆ xˆ
Reciprocal lattice vectors are
* a2 a3 3a 2 xˆ 4 4
a1 2 2 xˆ 0 yˆ xˆ 0 yˆ
V 3 3 3 3a 3
a
2
a3 a1
3 2
a
3 yˆ xˆ
2 2 2
*
a2 2
V
2 2
3 3 3
3a
xˆ 3 yˆ
3
xˆ
3
yˆ
a
2
* 4 2 2
for a 1 : a1 xˆ 0 yˆ , a2* xˆ yˆ
3 3 3
Q41. Consider a one-dimensional chain of atoms with lattice constant a . The energy of an
electron with wave-vector k is k cos ka , where and are constants. If
an electric field E is applied in the positive x -direction, the time dependent velocity of
an electron is
(In the following B is the constant)
eE
(a) Proportional to cos B at (b) proportional to E
eE
(c) independent of E (d) proportional to sin B at
Ans. : (d)
Solution: In the presence of electric field E , we can write
dp dk
F eE eE eE
dt dt
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eE
Integration gives, k t k 0 t
d 1 k
The group velocity v
dk dk
k
Since, k cos ka , a sin ka
k
a
Thus, v sin ka
Time dependent velocity of electron is
a a eE
v t sin k t a sin k 0 ta
a eE a eE
sin k 0 a at v t sin B at
Q42. A thin rectangular conducting plate of length a and width b is placed in the xy -plane in
two different orientations as shown in the figures below. In both cases a magnetic field B
is applied in the z -direction and a current flows in the x direction due to the applied
voltage V . b
y B
B
x a
V1 a
V1 b
V V
If the Hall voltage across the y -direction in the two cases satisfy V2 2V1 the ratio a : b
must be
(a) 1: 2 (b) 1: 2 (c) 2 :1 (d) 2 :1
Ans. : (d)
IB
Solution: Since, Hall voltage is given by VH , where w is width of conducting plate.
w
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l a
Since, in case (I), V I1 R1 and R1 1
A1 ab b
I1 bV
V I1
b
I1 B bVB bVB
Then, VH V1 w a
w 2w 2a
l2 b
And also in case (II), R2
A2 ab a
V Va
V I 2 R2 I 2
R2
I 2 B VaB
Then, VH V2
w 2b
a2 2
Since, V2 2V1 a : b 2 :1
b2 1
NET/JRF (JUNE-2017)
Q43. The energy gap and lattice constant of an indirect band gap semiconductor are 1.875 eV
and 0.52 nm , respectively. For simplicity take the dielectric constant of the material to be
unity. When it is excited by broadband radiation, an electron initially in the valence band
at k 0 makes a transition to the conduction band. The wavevector of the electron in the
conduction band, in terms of the wavevector kmax at the edge of the Brillouin zone, after
the transition is closest to
(a) kmax /10 (b) kmax /100 (c) kmax /1000 (d) 0
Ans. : (a)
Solution: The K value of electron in C.B. is
K K 7 109 m 1
1.05 1034 J .S .
2 2 3.14 K
K max at the Brillouin Zero is K max 9
1.2 1010 m 1 K max
a 0.52 10 m 10
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Q44. The electrical conductivity of copper is approximately 95% of the electrical conductivity
of silver, while the electron density in silver is approximately 70 % of the electron
density in copper. In Drude’s model, the approximate ratio Cu / Ag of the mean collision
cu 0.95 Ag 0.7ncu
0.66
Ag Ag ncu
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NUCLEAR AND PARTICLE PHYSICS
NET/JRF (JUNE-2011)
RCu ACu 64
RMg 3 3
RMg 4.8 10 13 3.6 10 13 cm.
RCu 4 4
(B). The root-mean-square (r.m.s) energy of a nucleon in a nucleus of atomic number A
in its ground state varies as:
(a) A4 / 3 (b) A1 / 3 (c) A1/ 3 (d) A2 / 3
Ans: (c)
Q2. A beam of pions (π+) is incident on a proton target, giving rise to the process
p n
(A). Assuming that the decay proceeds through strong interactions, the total isospin I and
its third component I3 for the decay products, are
3 3 5 5
(a) I , I 3 (b) I , I 3
2 2 2 2
5 3 1 1
(c) I , I 3 (d) I , I 3
2 2 2 2
Ans: (c)
1 5 1 3
Solution: p n ; I : 1 1 , I3 : 11
2 2 2 2
(B). Using isospin symmetry, the cross-section for the above process can be related to
that of the process
(a) n p (b) p n
(c) n p (d) p n
Ans: (c)
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NET/JRF (DEC-2011)
Q3. According to the shell model the spin and parity of the two nuclei 125
51 Sb and 89
38 Sr are,
respectively,
5 5 5 7
(a) and (b) and
2 2 2 2
7 5 7 7
(c) and (d) and
2 2 2 2
Ans: (d)
Solution: 125
51 Sb ; Z 51 and N 74
Z 51
7 7
j and l 4 . Thus spin and parity
2 2
89
38 Sr ; Z 38 and N 51
N 51:
7 7
j and l 4 . Thus spin and parity
2 2
Q4. The difference in the Coulomb energy between the mirror nuclei 49
24 Cr and 49
25 Mn is
6.0 MeV . Assuming that the nuclei have a spherically symmetric charge distribution and
5 6
25 2 24 2 4.9 10 15 m .
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NET/JRF (JUNE-2012)
1
Q5. The ground state of 12 Pb nucleus has spin-parity J , while the first excited state
207 p
2
5
has J p
.The electromagnetic radiation emitted when the nucleus makes a transition
2
from the first excited state to ground state are
(a) E2 and E3 (b) M2 or E3 (c) E2 or M3 (d) M2 or M3
Ans: (c)
Solution: No parity change; J 2,3
1 1
I 3 : 1 1 (Conserved)
2 2
(B) K K (Electromagnetic interaction)
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NET/JRF (JUNE-2013)
Q7. The binding energy of a light nucleus Z , A in MeV is given by the approximate formula
B A, Z 16 A 20 A 2/3 3
Z 2 A 1 / 3 30
N Z 2
4 A
where N A Z is the neutron number. The value of Z of the most stable isobar for a
given A is
1 1 1
A A2 / 3 A A A2 / 3 A A4 / 3
(a) 1 (b) (c) 1 (d) 1
2 160 2 2 120 2 64
Ans: (a)
1
B A A2 / 3
Solution: 0 Z 1
Z Z Z 2 160
6 10 3
1
5 t 1 N
t 2 10 sec . Thus e 2
1 e 1 .
3 10 8
2 N0
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NET/JRF -(DEC-2013)
A
Q10. The intrinsic electric dipole moment of a nucleus Z X
(a) increases with Z , but independent of A
(b) decreases with Z , but independent of A
(c) is always zero
(d) increases with Z and A
Ans: (d)
Q11. According to the shell model, the total angular momentum (in units of ) and the parity
of the ground state of the 37 Li nucleus is
3 3
(a) with negative parity (b) with positive parity
2 2
1 7
(c) with positive parity (d) with negative parity
2 2
Ans: (a)
Solution: Z 3, N 4
NET/JRF (JUNE-2014)
Q12. The recently-discovered Higgs boson at the LHC experiment has a decay mode into a
photon and a Z boson. If the rest masses of the Higgs and Z boson are 125 GeV/c 2 and
90 GeV/c 2 respectively, and the decaying Higgs particle is at rest, the energy of the
photon will approximately be
(a) 35 3 GeV (b) 35 GeV (c) 30 GeV (d) 15 GeV
Ans: (c)
Solution: Assume H is symbol of Higgs boson, H Z
E H2 E Z2 1252 902
E 30GeV
2EH 2 125
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Q13. In a classical model, a scalar (spin-0) meson consists of a quark and an antiquark bound
b
by a potential V r ar , where a 200 MeV fm -1 and b 100 MeV fm . If the
r
masses of the quark and antiquark are negligible, the mass of the meson can be estimated
as approximately
(a) 141 MeV/c 2 (b) 283 MeV/c 2 (c) 353 MeV/c 2 (d) 425 MeV/c 2
Ans: (b)
Solution: At equilibrium separation the potential is minimum, thus the equilibrium separation
can be determined as
dV r b b 100MeVfm 1
a 0 r0 1
fm
dr r r0
r02
a 200MeVfm 2
E 200 2 283MeV E m c 2
E
the mass of the meson m 283MeV / c 2
c 2
NET/JRF (DEC-2014)
Q14. Consider the four processes
(i) p n e ve (ii) 0 p e v e
(iii) e ve (iv) 0
which of the above is/are forbidden for free particles?
(a) only (ii) (b) (ii) and (iv) (c) (i) and (iv) (d) (i) and (ii)
Ans: (d)
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Solution: (i) p n e e [Not allowed]
It violate energy conservation. The mass of proton is less than mass of neutron. Free
proton is stable and can not decay to neutron. Proton can decay to neutron only inside the
nucleus, where energy violation is taken care by Heisenberg uncertainty principle.
(ii) 0 p e e [Not allowed]. In this decay charge is not conserved
h h
According to de-Broglie relation,
p 2mE
0 150
This can be also written as
E eV
150 150
E eV 1.04 1012 E 1.04 1012 eV
1.2 10
2 5 2
0
The bet suitable answer is option (b).
Q16. If the binding energy B of a nucleus (mass number A and charge Z ) is given by
2Z A
2
aC Z 2
B aV A aS A 2/3
asym 1/ 3
A A
where aV 16 MeV , a S 16 MeV , a sym 24 MeV and aC 0.75 MeV , then for the most
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dB 2 2Z A 2 2aC Z
Solution: For the most stable isobar for a nucleus 0 asym 1/ 3 0
dZ A A
2 2Z 216 2 2Z 4 2Z 216 3 2Z
24 0.75 0 0
216
1/ 3
216 9 4 6
4 2Z 216 Z
0 16 2 Z 216 9 Z 0 41Z 216 16 Z 82.3
9 4
NET/JRF (JUNE-2015)
Q17. The reaction 2
1 D 12 D 42 He 0 cannot proceed via strong interactions because it
violates the conservation of
(a) angular momentum (b) electric charge
(c) baryon number (d) isospin
Ans. (d)
Solution: 1 D 2 1 D 2 2 He 4 0 (Not conserved)
I: 0 0 0 1
This isopin is not conserved in above reaction.
Q18. Let us approximate the nuclear potential in the shell model by a three dimensional
isotropic harmonic oscillator. Since the lowest two energy levels have angular momenta
l 0 and l 1 respectively, which of the following two nuclei have magic numbers of
protons and neutrons?
(a) 42 He and 16
8 O (b) 12 D and 84 Be (c) 42 He and 84 Be (d) 42 He and 12
6 C
Ans. (a)
Solution: 2 He 4 has Z 2, N 2
Q19. The charm quark S assigned a charm quantum number C 1 . How should the
Gellmann-Nishijima formula for electric charge be modified for four flavors of quarks?
1 1
(a) I 3 B S C (b) I 3 B S C
2 2
1 1
(c) I 3 B S C (d) I 3 B S C
2 2
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Ans. (d)
1
Solution: From Gell-Mann-Nishijima formula Q I 3 B S
2
1
For Quark it is generalized as Q I 3 B S C
2
NET/JRF (DEC-2015)
Q20. Consider the following processes involving free particles
(i) n p e ve (ii) p n
(iii) p n 0 0 (iv) p ve n e
Which of the following statements is true?
(a) Process (i) obeys all conservation laws
(b) Process (ii) conserves baryon number, but violates energy-momentum conservation
(c) process (iii) is not allowed by strong interaction but is allowed by weak interactions
(d) Process (iv) conserves baryon number, but violates lepton number conservation
Ans.: (b)
Solution: (i) n p e ve
q 0 1 1 0 (conserved)
1 1 1 1
spin (not conserved)
2 2 2 2
Le 0 0 1 1 (not conserved)
(ii) Baryon number is conserved but energy and momentum conservation violated.
(iii) spin is not conserved
(iv) obeys all conservation laws.
Q21. Of the nuclei of mass number A 125 , the binding energy calculated from the liquid
drop model (given that the coefficients for the Coulomb and the asymmetry energy are
ac 0.7 MeV and asym 22.5 MeV respectively) is a maximum for
(a) 125
54 Xe (b) 124
53 I (c) 125
52 Te (d) 125
51 Sb
Ans.: (c)
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4 22.5 125 0.7 53
2/3
4aa ac A1/ 3 4aa A ac A2 / 3
Solution: Z 0 Z0
2ac A1/ 3 8aa A1 8aa 2ac A2 / 3 8 22.5 2 0.7 53
2/3
1 1
only N 0 atoms of X , at short times ( t as well as ) the number of atoms of Z
1 2
will be
1 12
(a) 12 N 0t 2 (b) N 0t
2 2 1 2
(c) 1 2 N 0t 2 (d) 1 2 N 0t
2
Ans: (a)
1 2
X Y Z
Solution: t 0 N0 0 0
t N1 N2 N3
dN dN
Rate equations N1 N 0 e 1t , 2 1 N1 2 N 2 , 3 2 N 2
dt dt
e 2t e 1 t
N 3 N 0 1 1 2
2 1 2 1
1 22t 2 2 12t 2
N 0 1 1 t 1 t
2 1 2 2 1
2 1
2
1 t 1 22t 2 2 2 1t 2 12t 2
N 0 1 1 2
2 1 2 1 2 1 2 2 1 2 1 2 1 2
1 2t 2 2 2t 2 t 2 1 1
N0 2 1 1 2 N0 2 12 N 0t
2
2 1
2 2 1
2 2
2 1 2 1 2
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Q23. In the large hadron collider LHC , two equal energy proton beams traverse in opposite
directions along a circular path of length 27 km . If the total centre of mass energy of a
proton-proton pair is 14 TeV , which of the following is the best approximation for the
proper time taken by a proton to traverse the entire path?
(a) 12 ns (b) 1.2 s (c) 1.2 ns (d) 0.12 s
Ans: (a)
Solution: The proton travel at nearly speed of light in LHC , therefore
d 27 103
t 9 105 sec
c 3 108
v2 t
Since, proton is relativistic, t0 t 1 2
c
1 m0 c 2 938 MeV 938 106 eV
E m0 c 2 1.34 104
E 7 TeV 7 1012 eV
t
Thus, t0 9 105 1.34 104 1.2 108 sec 12 ns
Q24. Let ES denotes the contribution of the surface energy per nucleon in the liquid drop
Q25. According to the shell model, the nuclear magnetic moment of the 27
13 Al nucleus is (Given
2 j 1 g S N 2 1 5.586 N 4.793 N
1 1 5
Magnetic moment,
2 2 2
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NET/JRF (DEC-2016)
Q26. What should be the minimum energy of a photon for it to split an -particle at rest into a
tritium and a proton?
(The masses of 4
2 He, 13 H and 1
1 H are 4.0026 amu,3.0161 amu and 1.0073 amu
Q27. Which of the following reaction(s) is/are allowed by the conservation laws?
(i) n 0 K
(ii) p 0 K 0
(a) both (i) and (ii) (b) only (i)
(c) only (ii) (d) neither (i) nor (ii)
Ans. : (a)
Solution: (i) n 0 K
q :1 0 0 1
B : 0 1 1 0
S : 0 0 1 1
Reaction is allowed
(ii) p 0 K 0
q : 1 1 0 0
B : 0 1 1 0
S : 0 0 1 1
Reaction is allowed
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Q28. A particle, which is a composite state of three quarks u , d and s , has electric charge,
spin and strangeness respectively, equal to
1 1 1
(a) 1, , 1 (b) 0, 0, 1 (c) 0, , 1 (d) 1, 1
2 2 2
Ans. : (c)
Solution: charge, spin and strangers of Quarks u , d & s are given as
U D S Total
Charge 2 1 1 0
3 3 3
Spin 1 1 1 1 3
or
2 2 2 2 2
Strangeness 0 0 1 1
If a particle x is a composite of u, d & s , then net charge, spin and strangeness on x is
net charge 0
1 3
net spin or and net strangeness 1
2 2
NET/JRF (JUNE-2017)
Q29. If in a spontaneous - decay of 232
92 U at rest, the total energy released in the reaction is
770 MeV / c 2 , then the range of the force due to exchange of rho-mesons is
(a) 1.40 fm (b) 7.70 fm (c) 0.25 fm (d) 0.18 fm
Ans. : (c)
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c
Solution: Range for nuclear force between nucleon will be R ct and c 199MeVfm
mc 2
199 MeVfm
R 0.25 fm
MeV 2
770 2 c
c
Q31. A baryon X decays by strong interaction as X 0 , where is a member
of the isotriplet , 0 , . The third component I 3 of the isospin of X is
I 3 :1
1 0
I 3 for X is 0 .
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